Give out candies
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Problem Description
There are n children numbered 1 to n, and HazelFan's task is to give out candies to the children. Each children must be given at least 1 and at most m candies. The children raised k pieces of requirements, and every piece includes three integers x,y,z, means that the number of candies given to the child numbered x, subtract the number of candies given to the child number y, the result should not be more than z. Besides, if the child numbered i is given j candies, he or she will get wi,jsatisfaction score, now you need to help HazelFan maximize the sum of the satisfaction scores of these children.
Input
The first line contains a positive integer T(1≤T≤5), denoting the number of test cases.
For each test case:
The first line contain three positive integers n,m,k(1≤n,m≤50,1≤k≤150).
The next n lines, the ith line contains m positive integers, the jth integer denotes wi,j.
The next k lines, each line contains three integers x,y,z(1≤x,y≤n,∣z∣<233), denoting a piece of requirement.
For each test case:
The first line contain three positive integers n,m,k(1≤n,m≤50,1≤k≤150).
The next n lines, the ith line contains m positive integers, the jth integer denotes wi,j.
The next k lines, each line contains three integers x,y,z(1≤x,y≤n,∣z∣<233), denoting a piece of requirement.
Output
For each test case:
A single line contains a nonnegative integer, denoting the answer. Particularly, if there is no way to give out candies, print -1.
A single line contains a nonnegative integer, denoting the answer. Particularly, if there is no way to give out candies, print -1.
Sample Input
2
2 1 1
1
1
1 2 1
3 3 2
1 2 3
2 3 1
3 1 2
1 2 0
2 3 0
Sample Output
2
7
题意：
有n个人，每个人i获得j个糖果会有一个满意度wij，现在有一些约数条件xi,yi,zi，即xi获得的糖果数量-yi获得的糖果数量不超过zi，求最大满意度。
题解：
用i.j这点表示给第i个孩子至少j块糖
当这个点属于S集合时所代表条件成立。
然后i.j->i.j+1连1000-wi,j，i.m向T连1000-wi,m，S向i.1连inf。
如果存在ax-ay>=z，x.k->y.k+z连inf。跑最小割，再用n*1000减掉答案。
#include<bits/stdc++.h> using namespace std; #define ls i<<1 #define rs ls | 1 #define mid ((ll+rr)>>1) #define pii pair<int,int> #define MP make_pair typedef long long LL; typedef unsigned long long ULL; const long long INF = 1e18+1LL; const double pi = acos(-1.0); const int N = 1e5+10, M = 1e3+20,inf = 1000000; int head[N],t=2,h[N],q[N],S,T,ans = 0,huyou[N]; struct edge{int to,next,v;}e[N * 2]; void adds(int u,int v,int w) {e[t].to=v;e[t].v=w;e[t].next=head[u];head[u]=t++;} void add(int u,int v,int w) {adds(u,v,w);adds(v,u,0);} inline int bfs() { for(int i = 0; i <= 5000; ++i) h[i] = -1; int l=0,r=1,now; q[l]=S; h[S]=0; while(l!=r){ now=q[l++];//if(l == 100000) l=0; for(int i=head[now];i!=-1;i=e[i].next) { if(e[i].v&&h[e[i].to]==-1) { h[e[i].to]=h[now]+1; q[r++]=e[i].to; // if(r==100000) r = 0; } } } if(h[T]==-1) return 0; else return 1; } inline int dfs(int x,int f) { if(x == T || f == 0) return f; int used=0,w; for(int &i=huyou[x]; i!=-1;i=e[i].next) { if(e[i].v&&h[e[i].to] == h[x] + 1) { w=dfs(e[i].to,min(f-used,e[i].v)); used+=w;e[i].v-=w;e[i^1].v+=w; f-=w; if(f == 0) break; } } return used; } void dinic() { while(bfs()) { memcpy(huyou,head,sizeof(huyou)); ans+=dfs(S,inf); } } int n,m,k,a[56][56]; int main() { int TT; scanf("%d",&TT); while(TT--) { scanf("%d%d%d",&n,&m,&k); S = 0,T = n*m+1; t = 2; for(int i = 0; i <= 50000; ++i) head[i] = -1; for(int i = 1; i <= n; ++i) { for(int j = 1; j <= m; ++j) scanf("%d",&a[i][j]); add(S,(i-1)*m+1,1000000); for(int j = 1; j < m; ++j) add((i-1)*m+j,(i-1)*m+j+1,10000-a[i][j]); add(i*m,T,10000-a[i][m]); } for(int i = 1; i <= k; ++i) { int x,y,z,last = 0; scanf("%d%d%d",&x,&y,&z); for(int j = max(z+1,1); j <= m && j-z <= m; ++j) { add((x-1)*m+j,(y-1)*m+j-z,1000000); last = j; } if(last + 1 - z > m && last+1<=m) add((x-1)*m + last+1,T,1000000); } ans = 0; dinic(); if(ans >= 1000000) puts("-1"); else printf("%d ",n*10000 - ans); } return 0; }