2053: Birthday
样例输入
3 3
1 2
1 2
1 2
样例输出
5
分析:
考虑费用流时把每个part拆成n个点,选择第i个点的代表为放置i块蛋糕和(i - 1)块蛋糕的时间差,这个时间差是递增的,因此在费用流的过程中必定会从小到大选择
具体建图:左边n个点代表n个蛋糕,右边m * n个点代表m个part,每个part拆成n个点。源点向每个左边的点连一条流量1费用0的边,每个右边的点向汇点连一条流量1费用0的编。每个蛋糕向可以放的两个part的所有点连边,连向第i个点的费用为i^2 - (i - 1)^2,流量为1。这样求最小费用流既为答案。
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 5005;
struct Edge{
int value,flow,to,rev;
Edge(){}
Edge(int a,int b,int c,int d):to(a),value(b),flow(c),rev(d){}
};
vector<Edge> E[MAXN];
inline void Add(int from,int to,int flow,int value){
E[from].push_back(Edge(to,value,flow,E[to].size()));
E[to].push_back(Edge(from,-value,0,E[from].size()-1));
}
bool book[MAXN];//用于SPFA中标记是否在queue中
int cost[MAXN];//存费用的最短路径
int pre[MAXN];//存前节点
int pree[MAXN];//存在前节点的vector中的下标
bool Spfa(int from,int to){
memset(book,false,sizeof book);
memset(cost,INF,sizeof cost);
book[from] = true;
cost[from] = 0;
queue<int> Q;
Q.push(from);
while(!Q.empty()){
int t = Q.front();
book[t] = false;
Q.pop();
for(int i=0 ; i<E[t].size() ; ++i){
Edge& e = E[t][i];
if(e.flow > 0 && cost[e.to] > cost[t] + e.value){
cost[e.to] = cost[t] + e.value;
pre[e.to] = t;
pree[e.to] = i;
if(book[e.to] == false){
Q.push(e.to);
book[e.to] = true;
}
}
}
}
return cost[to] != INF;
}
int Work(int from,int to){
int sum = 0;
while(Spfa(from,to)){
int mflow = INF;//SPFA找到的最短路径的最小容量
int flag = to;
while(flag != from){
mflow = min(mflow,E[pre[flag]][pree[flag]].flow);
flag = pre[flag];
}
flag = to;
while(flag != from){
sum += E[pre[flag]][pree[flag]].value * mflow;
E[pre[flag]][pree[flag]].flow -= mflow;
E[flag][E[pre[flag]][pree[flag]].rev].flow += mflow;
flag = pre[flag];
}
}
return sum;
}
int main(){
int N,M;
while(scanf("%d %d",&N,&M) == 2){
int a,b;
for(int i=1 ; i<=N ; ++i){
scanf("%d %d",&a,&b);
Add(i,a+N,1,0);
Add(i,b+N,1,0);
Add(0,i,1,0);
}
for(int i=1 ; i<=M ; ++i){
for(int j=1 ; j<=99 ; j+=2){//99 = 2*50-1;
Add(i+N,M+N+1,1,j);
}
}
printf("%d\n",Work(0,M+N+1));
for(int i=0 ; i<=M+N+1 ; ++i)E[i].clear();
}
return 0;
}
2088: 电音之王
样例输入
1
1 1 1 1 0 1000000007 10
样例输出
904493530
分析:
由于取模的常数为64bit整数,所以考虑使用O(1)快速乘,然后暴力求解即可
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll ksc(ll x,ll y,ll mod)
{
return (x*y-(ll)((long double)x/mod*y)*mod+mod)%mod;
}
ll m0,m1,c,M,k;
ll a0,a1;
int main()
{
int T;
scanf("%d",&T);
while(T--){
scanf("%lld%lld%lld%lld%lld%lld%lld",&a0,&a1,&m0,&m1,&c,&M,&k);
ll ans=ksc(a0,a1,M);
ll a2;
for(int i=2;i<=k;i++)
{
a2=((ksc(m0,a1,M)+ksc(m1,a0,M))%M+c)%M;
ans=ksc(ans,a2,M);
a0=a1;
a1=a2;
}
printf("%lld\n",ans);
}
return 0;
}