牛客网暑期ACM多校训练营（第二场）A run D money H travel I car J farm

其他 2018-09-10 14:08:56 阅读次数: 0
版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/Lfhase/article/details/81157619

比赛（题目）链接：https://www.nowcoder.com/acm/contest/140#question

A

签到题，每个点只会影响后面1步和k步的点，注意不要重复两次k就好了

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
 
using namespace std;
 
const int MAXN = 100005;
const int MOD = 1e9+7;
 
long long DP_1[MAXN];
long long DP_2[MAXN];
long long DP[MAXN];
long long STEP;
 
 
int main()
{
    int Q;
    while(scanf("%d%lld",&Q,&STEP)!=EOF)
    {
        memset(DP_1,0,sizeof(DP_1));
        memset(DP_2,0,sizeof(DP_2));
        //for(int i=1;i<MAXN;++i)    DP[i] = 1;
        DP_1[0] = 1;
        for(int i=0;i<MAXN;++i)
        {
            if(i+STEP<MAXN)
            {
                DP_2[STEP+i] = (DP_2[STEP+i]+DP_1[i])%MOD;
            }
            if(i+1<MAXN)
            {
                DP_1[i+1] = (DP_1[i+1]+DP_1[i]+DP_2[i])%MOD;
            }
        }
 
       /* for(int i=1;i<MAXN;++i)
            cout<<(DP_1[i]+DP_2[i])<<" ";
        cout<<endl;*/
        DP[0] = 0;
        for(int i=1;i<MAXN;++i)
        {
            DP[i] = (DP[i-1]+DP_1[i]+DP_2[i])%MOD;
            /*if(i<=100)
            cout<<DP[i]<<" ";*/
        }
 
 
        while(Q--)
        {
            long long res = 0;
            int l, r;
            scanf("%d%d",&l,&r);
            res = (DP[r]-DP[l-1])%MOD;
            if(res<0)   res+=MOD;
            printf("%lld\n",res);
        }
    }
    return 0;
}

D

签到题，贪心地取所有递增连续子序列最后两值相减即可。

AC代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
 
using namespace std;
 
const int MAXN = 100005;
const int MOD = 1e9+7;
 
int N;
 
 
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        cin>>N;
        long long profit = 0;
        long long times = 0;
        long long cur = -1;
        long long curnum;
        long long lastnum=-1;
        for(int i=1;i<=N;++i)
        {
            scanf("%lld",&curnum);
            if((lastnum!=-1&&cur==-1&&lastnum<curnum))
            {
                cur = lastnum;
            }
            else if(lastnum!=-1&&cur!=-1&&curnum<lastnum)
            {
                profit += lastnum-cur;
                times+=2;
                cur = -1;
            }
            lastnum = curnum;
        }
        if(cur!=-1&&lastnum>cur)    profit+=lastnum-cur, times+=2;
        printf("%lld %lld\n",profit,times);
    }
    return 0;
}

H 

树形DP

设置两个数组：

DP [ i, j ] ：表示树上前i个点选择j条链的最大值

SLink [ i, j ] : 表示树上前i个点选择j条链和一条包含i的直链（i为端点到叶子）的最大值

对于每一个子树，它对结果的贡献可以是：

1） None；

2）1，2，3条路径；

3） 一条直链+0，1，2，3条路径；

合并时按上述情况分类讨论即可，注意在存在直链的子树合并时，需要减1。

AC代码如下：

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>

using namespace std;

#define FSIO  ios::sync_with_stdio(0);cin.tie(0);
#define DEBUG(a)   cout<<"DEBUG: "<<(a)<<endl;

const int MAXN = 400005;
const int MOD = 1e9+7;
const int INF = 1e9+7;

int N;
int weight[MAXN];
vector<int> mapp[MAXN];

long long DP[MAXN][5];
long long SLink[MAXN][5];

void rua(int u, int fa=0)
{
    long long maxn[5][5], curmaxn[5][5], submaxn[5][5];
    memset(maxn,0,sizeof(maxn));
    memset(curmaxn,0,sizeof(curmaxn));
    memset(submaxn,0,sizeof(submaxn));
    for(int e:mapp[u])
    {
        if(e!=fa)
        {
            rua(e,u);
            for(int i=0;i<=3;++i)   submaxn[i][0]=DP[e][i], submaxn[i][1]=SLink[e][i];
            for(int x1=0;x1<=3;++x1)
                for(int x2=0;x2+x1<=3;++x2)
                    for(int sy1=0;sy1<=2;++sy1)
                        for(int sy2=0;sy2+sy1<=3;++sy2)
                            maxn[x1+x2][sy1+sy2] = max(maxn[x1+x2][sy1+sy2],curmaxn[x1][sy1]+submaxn[x2][sy2]);
            memcpy(curmaxn,maxn,sizeof(maxn));
        }
    }
    for(int i=0;i<=3;++i)   DP[u][i] = max(DP[u][i],maxn[i][0]);
    for(int i=1;i<=3;++i)   DP[u][i] = max(DP[u][i],maxn[i-1][1]+weight[u]);
    for(int i=1;i<=3;++i)   DP[u][i] = max(DP[u][i],maxn[i-1][2]+weight[u]);
    for(int i=0;i<=3;++i)   SLink[u][i] = max(SLink[u][i],maxn[i][1]+weight[u]);
}



int main()
{
    //freopen("output.txt","w+",stdout);
    int u, v;
    scanf("%d",&N);
    for(int i=1;i<=N;++i)   scanf("%d",weight+i);
    for(int i=1;i<N;++i)
    {
        scanf("%d%d",&u,&v);
        mapp[u].push_back(v);
        mapp[v].push_back(u);
    }

    rua(1,0);
    printf("%lld",DP[1][3]);

    return 0;
}

I

找规律？规律就是n为偶数时能通过的车为2n，n为奇数时能通过的车是2n-1

摧毁格子时对应行列减一，注意奇数情况下对中心行列的处理即可。

AC代码如下：

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>

using namespace std;

#define FSIO  ios::sync_with_stdio(0);cin.tie(0);
#define DEBUG(a)   cout<<"DEBUG: "<<(a)<<endl;

const int MAXN = 100005;
const int MOD = 1e9+7;
const int INF = 1e9+7;

int N, M;
int cordx[MAXN], cordy[MAXN];


int main()
{
    //freopen("output.txt","w+",stdout);
    while(scanf("%d%d",&N,&M)!=EOF)
    {
        int res = N*2-(N%2);
        memset(cordx,0,sizeof(cordx));
        memset(cordy,0,sizeof(cordy));
        int x, y;

        /* for(int i=1;i<=N;++i)
             cout<<cordx[i]<<" ";
         cout<<endl;
         for(int j=1;j<=N;++j)
             cout<<cordy[j]<<" ";
         cout<<endl;*/
        for(int i=1; i<=M; ++i)
        {
            scanf("%d%d",&x,&y);

            if(!cordx[x])
            {
                res --;
                cordx[x] = 1;
            }
            if(!cordy[y])
            {
                res--;
                cordy[y] = 1;
            }

        }
        if(cordx[(N+1)/2]==cordy[(N+1)/2]&&cordy[(N+1)/2]==1)   res++;
        /*for(int i=1;i<=N;++i)
            cout<<cordx[i]<<" ";
        cout<<endl;
        for(int j=1;j<=N;++j)
            cout<<cordy[j]<<" ";
        cout<<endl;*/

        printf("%d\n",res);
    }
    return 0;
}

J

二维树状数组

考虑到每种肥料只会对对应种类的植物有作用，所以按植物/肥料种类进行分类判断

对每种植物，对农场矩阵施加非该种植物的化肥，

然后对该种植物每个点做一次前缀和判断，如果是0则无影响，如果是1则有影响。

AC代码如下：（事实上不用IO加速也能过【tao

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>

using namespace std;

#define FSIO  ios::sync_with_stdio(0);cin.tie(0);
#define DEBUG(a)   cout<<"DEBUG: "<<(a)<<endl;
#define R register

namespace fastIO {
#define BUF_SIZE 100005
#define OUT_SIZE 100005
#define ll long long
  //fread->read
  bool IOerror = 0;
  inline char nc() {
    static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;
    if (p1 == pend) {
      p1 = buf;
      pend = buf + fread(buf, 1, BUF_SIZE, stdin);
      if (pend == p1) {
    IOerror = 1;
    return -1;
      }
      //{printf("IO error!\n");system("pause");for (;;);exit(0);}
    }
    return *p1++;
  }
  inline bool blank(char ch) {
    return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
  }
  inline void read(int &x) {
    bool sign = 0;
    char ch = nc();
    x = 0;
    for (; blank(ch); ch = nc());
    if (IOerror)return;
    if (ch == '-')sign = 1, ch = nc();
    for (; ch >= '0' && ch <= '9'; ch = nc())x = x * 10 + ch - '0';
    if (sign)x = -x;
  }
  inline void read(ll &x) {
    bool sign = 0;
    char ch = nc();
    x = 0;
    for (; blank(ch); ch = nc());
    if (IOerror)return;
    if (ch == '-')sign = 1, ch = nc();
    for (; ch >= '0' && ch <= '9'; ch = nc())x = x * 10 + ch - '0';
    if (sign)x = -x;
  }
  inline void read(double &x) {
    bool sign = 0;
    char ch = nc();
    x = 0;
    for (; blank(ch); ch = nc());
    if (IOerror)return;
    if (ch == '-')sign = 1, ch = nc();
    for (; ch >= '0' && ch <= '9'; ch = nc())x = x * 10 + ch - '0';
    if (ch == '.') {
      double tmp = 1;
      ch = nc();
      for (; ch >= '0' && ch <= '9'; ch = nc())tmp /= 10.0, x += tmp * (ch - '0');
    }
    if (sign)x = -x;
  }
  inline void read(char *s) {
    char ch = nc();
    for (; blank(ch); ch = nc());
    if (IOerror)return;
    for (; !blank(ch) && !IOerror; ch = nc()) * s++ = ch;
    *s = 0;
  }
  inline void read(char &c) {
    for (c = nc(); blank(c); c = nc());
    if (IOerror) {
      c = -1;
      return;
    }
  }
  //getchar->read
  inline void read1(int &x) {
    char ch;
    int bo = 0;
    x = 0;
    for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())if (ch == '-')bo = 1;
    for (; ch >= '0' && ch <= '9'; x = x * 10 + ch - '0', ch = getchar());
    if (bo)x = -x;
  }
  inline void read1(ll &x) {
    char ch;
    int bo = 0;
    x = 0;
    for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())if (ch == '-')bo = 1;
    for (; ch >= '0' && ch <= '9'; x = x * 10 + ch - '0', ch = getchar());
    if (bo)x = -x;
  }
  inline void read1(double &x) {
    char ch;
    int bo = 0;
    x = 0;
    for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())if (ch == '-')bo = 1;
    for (; ch >= '0' && ch <= '9'; x = x * 10 + ch - '0', ch = getchar());
    if (ch == '.') {
      double tmp = 1;
      for (ch = getchar(); ch >= '0' && ch <= '9'; tmp /= 10.0, x += tmp * (ch - '0'), ch = getchar());
    }
    if (bo)x = -x;
  }
  inline void read1(char *s) {
    char ch = getchar();
    for (; blank(ch); ch = getchar());
    for (; !blank(ch); ch = getchar()) * s++ = ch;
    *s = 0;
  }
  inline void read1(char &c) {
    for (c = getchar(); blank(c); c = getchar());
  }
  //scanf->read
  inline void read2(int &x) {
    scanf("%d", &x);
  }
  inline void read2(ll &x) {
#ifdef _WIN32
    scanf("%I64d", &x);
#else
#ifdef __linux
    scanf("%lld", &x);
#else
    puts("error:can recognize the system!");
#endif
#endif
  }
  //fwrite->write
  struct Ostream_fwrite {
    char *buf, *p1, *pend;
    Ostream_fwrite() {
      buf = new char[BUF_SIZE];
      p1 = buf;
      pend = buf + BUF_SIZE;
    }
    void out(char ch) {
      if (p1 == pend) {
    fwrite(buf, 1, BUF_SIZE, stdout);
    p1 = buf;
      }
      *p1++ = ch;
    }
    void print(int x) {
      static char s[15], *s1;
      s1 = s;
      if (!x)*s1++ = '0';
      if (x < 0)out('-'), x = -x;
      while(x)*s1++ = x % 10 + '0', x /= 10;
      while(s1-- != s)out(*s1);
    }
    void println(int x) {
      static char s[15], *s1;
      s1 = s;
      if (!x)*s1++ = '0';
      if (x < 0)out('-'), x = -x;
      while(x)*s1++ = x % 10 + '0', x /= 10;
      while(s1-- != s)out(*s1);
      out('\n');
    }
    void print(ll x) {
      static char s[25], *s1;
      s1 = s;
      if (!x)*s1++ = '0';
      if (x < 0)out('-'), x = -x;
      while(x)*s1++ = x % 10 + '0', x /= 10;
      while(s1-- != s)out(*s1);
    }
    void println(ll x) {
      static char s[25], *s1;
      s1 = s;
      if (!x)*s1++ = '0';
      if (x < 0)out('-'), x = -x;
      while(x)*s1++ = x % 10 + '0', x /= 10;
      while(s1-- != s)out(*s1);
      out('\n');
    }
    void print(double x, int y) {
      static ll mul[] = {1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000,
            1000000000, 10000000000LL, 100000000000LL, 1000000000000LL, 10000000000000LL,
            100000000000000LL, 1000000000000000LL, 10000000000000000LL, 100000000000000000LL
      };
      if (x < -1e-12)out('-'), x = -x;
      x *= mul[y];
      ll x1 = (ll)floor(x);
      if (x - floor(x) >= 0.5)++x1;
      ll x2 = x1 / mul[y], x3 = x1 - x2 * mul[y];
      print(x2);
      if (y > 0) {
    out('.');
    for (size_t i = 1; i < y && x3 * mul[i] < mul[y]; out('0'), ++i);
    print(x3);
      }
    }
    void println(double x, int y) {
      print(x, y);
      out('\n');
    }
    void print(char *s) {
      while (*s)out(*s++);
    }
    void println(char *s) {
      while (*s)out(*s++);
      out('\n');
    }
    void flush() {
      if (p1 != buf) {
    fwrite(buf, 1, p1 - buf, stdout);
    p1 = buf;
      }
    }
    ~Ostream_fwrite() {
      flush();
    }
  } Ostream;
  inline void print(int x) {
    Ostream.print(x);
  }
  inline void println(int x) {
    Ostream.println(x);
  }
  inline void print(char x) {
    Ostream.out(x);
  }
  inline void println(char x) {
    Ostream.out(x);
    Ostream.out('\n');
  }
  inline void print(ll x) {
    Ostream.print(x);
  }
  inline void println(ll x) {
    Ostream.println(x);
  }
  inline void print(double x, int y) {
    Ostream.print(x, y);
  }
  inline void println(double x, int y) {
    Ostream.println(x, y);
  }
  inline void print(char *s) {
    Ostream.print(s);
  }
  inline void println(char *s) {
    Ostream.println(s);
  }
  inline void println() {
    Ostream.out('\n');
  }
  inline void flush() {
    Ostream.flush();
  }
  //puts->write
  char Out[OUT_SIZE], *o = Out;
  inline void print1(int x) {
    static char buf[15];
    char *p1 = buf;
    if (!x)*p1++ = '0';
    if (x < 0)*o++ = '-', x = -x;
    while(x)*p1++ = x % 10 + '0', x /= 10;
    while(p1-- != buf)*o++ = *p1;
  }
  inline void println1(int x) {
    print1(x);
    *o++ = '\n';
  }
  inline void print1(ll x) {
    static char buf[25];
    char *p1 = buf;
    if (!x)*p1++ = '0';
    if (x < 0)*o++ = '-', x = -x;
    while(x)*p1++ = x % 10 + '0', x /= 10;
    while(p1-- != buf)*o++ = *p1;
  }
  inline void println1(ll x) {
    print1(x);
    *o++ = '\n';
  }
  inline void print1(char c) {
    *o++ = c;
  }
  inline void println1(char c) {
    *o++ = c;
    *o++ = '\n';
  }
  inline void print1(char *s) {
    while (*s)*o++ = *s++;
  }
  inline void println1(char *s) {
    print1(s);
    *o++ = '\n';
  }
  inline void println1() {
    *o++ = '\n';
  }
  inline void flush1() {
    if (o != Out) {
      if (*(o - 1) == '\n')*--o = 0;
      puts(Out);
    }
  }
  struct puts_write {
    ~puts_write() {
      flush1();
    }
  } _puts;
  inline void print2(int x) {
    printf("%d", x);
  }
  inline void println2(int x) {
    printf("%d\n", x);
  }
  inline void print2(char x) {
    printf("%c", x);
  }
  inline void println2(char x) {
    printf("%c\n", x);
  }
  inline void print2(ll x) {
#ifdef _WIN32
    printf("%I64d", x);
#else
#ifdef __linux
    printf("%lld", x);
#else
    puts("error:can recognize the system!");
#endif
#endif
  }
  inline void println2(ll x) {
    print2(x);
    printf("\n");
  }
  inline void println2() {
    printf("\n");
  }
#undef ll
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace fastIO;


const int MAXN = 1000005;
const int MOD = 1e9+7;
const int INF = 1e9+7;

int N, M;

//  2-d BIT
vector<int> data[MAXN];


void Add(int x, int y, int z) {
    for (int i = x; i <= N; i += i & -i)
        for (int j = y; j <= M; j += j & -j)
            data[i][j] += z;
}
int Sum(int x, int y) {
    int ans = 0;
    for (int i = x; i; i -= i & -i)
        for (int j = y; j; j -= j & -j)
            ans += data[i][j];
    return ans;
}

// end of 2-d BIT

vector<pair<int,int> > plants[MAXN];
vector<pair<pair<int,int>,pair<int,int>>> fertilizer[MAXN];

int main()
{
    //freopen("output.txt","w+",stdout);
    int T;
    //scanf("%d%d%d",&N,&M,&T);
    read(N);
    read(M);
    read(T);
    for(int i=1; i<=N+1; ++i)
        for(int j=1; j<=M+1; ++j)
            data[i].push_back(0);
    int plantype;
    for(int i=1; i<=N; ++i)
        for(int j=1; j<=M; ++j)
            {
                //scanf("%d",&plantype);
                read(plantype);
                plants[plantype].push_back({i,j});
            }
    int x1, y1, x2, y2, ftype;
    for(int i=1;i<=T;++i)
    {
        //scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&ftype);
        read(x1); read(y1); read(x2); read(y2); read(ftype);
        fertilizer[ftype].push_back({{x1,y1},{x2,y2}});
        Add(x1,y1,1);
        Add(x2+1,y1,-1);
        Add(x1,y2+1,-1);
        Add(x2+1,y2+1,1);
    }
    int res = 0;
    for(int i=1;i<=N*M;++i)
    {
        for(int j=0;j<fertilizer[i].size();++j)
        {
            Add(fertilizer[i][j].first.first,fertilizer[i][j].first.second,-1);
            Add(fertilizer[i][j].second.first+1,fertilizer[i][j].first.second,1);
            Add(fertilizer[i][j].first.first,fertilizer[i][j].second.second+1,1);
            Add(fertilizer[i][j].second.first+1,fertilizer[i][j].second.second+1,-1);
        }
        for(int j=0;j<plants[i].size();++j)
            if(Sum(plants[i][j].first,plants[i][j].second))
                res++;
        for(int j=0;j<fertilizer[i].size();++j)
        {
            Add(fertilizer[i][j].first.first,fertilizer[i][j].first.second,1);
            Add(fertilizer[i][j].second.first+1,fertilizer[i][j].first.second,-1);
            Add(fertilizer[i][j].first.first,fertilizer[i][j].second.second+1,-1);
            Add(fertilizer[i][j].second.first+1,fertilizer[i][j].second.second+1,1);
        }
    }
    //printf("%d\n",res);
    println(res);
    return 0;
}

猜你喜欢

转载自blog.csdn.net/Lfhase/article/details/81157619
今日推荐
TIOBE 5 月榜单：Fortran “复活”进入 Top 10
GCC 14.1 发布
面壁智能发布 Eurux-8x22B 开源大模型 —— 堪称「理科状元」
开源日报 | 谷歌扶持鸿蒙上位；开源Rabbit R1；Docker加持的安卓手机；微软的焦虑和野心；海尔电器把开放平台关了
中国码农的“35岁魔咒”
蘭雅 CorelDRAW 插件 2024.5.1 国际劳动节版，免费下载
Arc Browser for Windows 1.0 正式 GA
90后程序员开发视频搬运软件、不到一年获利超 700 万，结局很刑！
周排行
Java自定义时间格式
同步整形电路
在开发中最最最常用的字符串的属性大集合
Linux 查看端口占用并杀掉
Java基础四：ArrayList
多线程之死锁就是这么简单
mysql 基础命令集
awk 命令详解
Centos6.3编译安装nginx+php步骤
OCR （Optical Character Recognition，光学字符识别）
每日归档
更多
2024-05-08(42)
2024-05-07(14)
2024-05-06(40)
2024-05-05(0)
2024-05-04(7)
2024-05-03(19)
2024-05-02(0)
2024-05-01(4)
2024-04-30(1)
2024-04-29(40)

代码天地 © 2018 codetd.com

境外服务器   最新电视剧2022