一共有 X + Y + Z X+Y+Z X+Y+Z 个人,每个人有 A i A_i Ai 个金钢镚, B i B_i Bi 个银钢镚, C i C_i Ci 个铜钢镚,你现在要从 X X X 个人那里拿走他们的金钢镚,从 Y Y Y 个人那里拿走他们的银钢镚,从 Z Z Z 个人那里拿走他们的铜钢镚。每个人被收且只被收了一次钢镚,那么你最多能拿到多少钢镚呢?
第一行三个正整数,分别为 X , Y , Z X,Y,Z X,Y,Z
之后 X + Y + Z X+Y+Z X+Y+Z 行,每行三个非负整数 A i , B i , C i A_i,B_i,C_i Ai,Bi,Ci
一行,表示最多能拿到的钢镚数目
样例输入1
1 2 1
2 4 4
3 2 1
7 6 7
5 2 3
样例输出1
18
样例输入2
6 2 4
33189 87907 277349742
71616 46764 575306520
8801 53151 327161251
58589 4337 796697686
66854 17565 289910583
50598 35195 478112689
13919 88414 103962455
7953 69657 699253752
44255 98144 468443709
2332 42580 752437097
39752 19060 845062869
60126 74101 382963164
样例输出2
3093929975
A i , B i , C i ≤ 1 0 9 A_i,B_i,C_i \le 10^9 Ai,Bi,Ci≤109
编号 | X + Y + Z ≤ X+Y+Z \le X+Y+Z≤ | 特殊性质 |
---|---|---|
1 − 4 1-4 1−4 | 200000 200000 200000 | Z = 0 Z=0 Z=0 |
5 − 8 5-8 5−8 | 1000 1000 1000 | |
9 − 20 9-20 9−20 | 200000 200000 200000 |
noi2019 集训 5.6
①模拟费用流
#include<bits/stdc++.h>
#define N 200005
typedef long long ll;
using namespace std;
const ll inf=1e15;
inline int read(){
int x=0,f=1;char s=getchar();
while(s<'0'||s>'9'){
if(s=='-')f=-1;s=getchar();}
while(s>='0'&&s<='9'){
x=(x<<3)+(x<<1)+s-'0';s=getchar();}
return x*f;
}
ll w[N][4],li[4],be[N];
priority_queue<pair<ll, int> > q[4][4];
int sz[4];
ll ans;
inline void ins(int x,int j){
sz[j]++;be[x]=j;
for(int k=1;k<=3;++k)if(k!=j){
q[j][k].push(make_pair(w[x][k]-w[x][j],x));
}
}
inline void del(int x,int j,int k){
sz[j]--;be[x]=0;
ins(x,k);
}
int main(){
// freopen("gangbeng.in","r",stdin);
// freopen("gangbeng.out","w",stdout);
// freopen("data.in","r",stdin);
// freopen("gangbeng.out","w",stdout);
int x=read(),y=read(),z=read(),n=x+y+z;
li[1]=x,li[2]=y,li[3]=z;
for(int i=1;i<=n;++i)for(int j=1;j<=3;++j)w[i][j]=read();
for(int i=1;i<=n;++i){
ll maxn=-inf,now[4]={
-inf,-inf,-inf,-inf};int res=0,flag=1;
for(int j=1;j<=3;++j){
for(int k=1;k<=3;++k)if(k!=j){
while(!q[j][k].empty()&&be[q[j][k].top().second]!=j)q[j][k].pop();
}
}
//cout<<1<<endl;
//cout<<i<<" "<<sz[1]<<" "<<sz[2]<<" "<<sz[3]<<endl;
for(int j=1;j<=3;++j){
if(!li[j])continue;
if(sz[j]<li[j]){
now[j]=w[i][j];
}else{
for(int k=1;k<=3;++k){
if(!li[k])continue;
if(k==j)continue;
res=1;while(res==j||res==k)++res;
if(sz[k]<li[k]){
//if(q[j][k].empty())cout<<i<<" "<<j<<" "<<k<<endl;
now[j]=max(now[j],w[i][j]+q[j][k].top().first);
}else{
//if(q[k][res].empty())cout<<i<<" "<<j<<" "<<k<<" "<<res<<endl;
if(li[res]){
now[j]=max(now[j],w[i][j]+q[j][k].top().first+q[k][res].top().first);
}
}
}
}
maxn=max(maxn,now[j]);
}
now[1]=now[2]=now[3]=-inf;
for(int j=1;j<=3;++j){
if(!li[j])continue;
if(flag==0)break;
if(sz[j]<li[j]){
now[j]=w[i][j];
if(now[j]==maxn&&flag){
flag=0;ins(i,j);
}
}else{
for(int k=1;k<=3;++k){
if(!li[k])continue;
if(k==j)continue;
res=1;while(res==j||res==k)++res;
if(sz[k]<li[k]){
now[j]=max(now[j],w[i][j]+q[j][k].top().first);
if(now[j]==maxn&&flag){
flag=0;del(q[j][k].top().second,j,k);
ins(i,j);
}
}else{
if(li[res]){
now[j]=max(now[j],w[i][j]+q[j][k].top().first+q[k][res].top().first);
if(now[j]==maxn&&flag){
flag=0;int x=q[j][k].top().second,y=q[k][res].top().second;
del(x,j,k);del(y,k,res);ins(i,j);
}
}
}
}
}
}
ans=ans+maxn;
}
printf("%lld\n",ans);
return 0;
}/*
1 2 1
2 4 4
3 2 1
7 6 7
5 2 3
*/
②简单贪心
#include<algorithm>
#include<cstdio>
#include<queue>
using namespace std;
struct ppap
{
long long a,b,c;
bool operator < (const ppap &A) const
{
return a-b<A.a-A.b;
}
}b[100010];
priority_queue<long long,vector<long long>,greater<long long> > q;
long long x,y,z,n,S,sum,ans,s[100010];
int main()
{
scanf("%lld%lld%lld",&x,&y,&z),n=x+y+z;
for(int i=1;i<=n;i++) scanf("%lld%lld%lld",&b[i].a,&b[i].b,&b[i].c),S+=b[i].c;
sort(b+1,b+1+n);
for(int i=1;i<=y;i++) sum+=b[i].b-b[i].c,q.push(b[i].b-b[i].c);
for(int i=y+1;i<=n-x+1;i++) s[i]=sum,sum+=b[i].b-b[i].c,q.push(b[i].b-b[i].c),sum-=q.top(),q.pop();
while(!q.empty()) q.pop();
sum=0;
for(int i=n;i>n-x;i--) sum+=b[i].a-b[i].c,q.push(b[i].a-b[i].c);
for(int i=n-x;i>=y;i--) ans=max(ans,S+s[i+1]+sum),sum+=b[i].a-b[i].c,q.push(b[i].a-b[i].c),sum-=q.top(),q.pop();
printf("%lld",ans);
}