HDU 6126.Give out candies 最小割

Give out candies

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 218    Accepted Submission(s): 66

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.

 

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.

 

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

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6126

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。

思路：最小割。应为求的是最大满意度，说以先要化为最小。约数条件就建立inf边，这样就不会进行穿过inf边的分割，约数条件也就成立了。具体建图看代码。

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
#define PI acos(-1.0)
const int maxn=1e5+100,maxm=1e5+100,inf=0x3f3f3f3f,mod=1e9+7;
const ll INF=1e18+7;
struct edge
{
    int from,to,cap,flow;
};
vector<edge>es;
vector<int>G[maxn];
bool vis[maxn];
int dist[maxn];
int iter[maxn];
void init(int n)
{
    for(int i=0; i<=n+10; i++) G[i].clear();
    es.clear();
}
void addedge(int from,int to,int cap)
{
    es.push_back((edge)
    {
        from,to,cap,0
    });
    es.push_back((edge)
    {
        to,from,0,0
    });
    int x=es.size();
    G[from].push_back(x-2);
    G[to].push_back(x-1);
}
bool BFS(int s,int t)
{
    memset(vis,0,sizeof(vis));
    queue <int> Q;
    vis[s]=1;
    dist[s]=0;
    Q.push(s);
    while(!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        for (int i=0; i<G[u].size(); i++)
        {
            edge &e=es[G[u][i]];
            if (!vis[e.to]&&e.cap>e.flow)
            {
                vis[e.to]=1;
                dist[e.to]=dist[u]+1;
                Q.push(e.to);
            }
        }
    }
    return vis[t];
}
int DFS(int u,int t,int f)
{
    if(u==t||f==0) return f;
    int flow=0,d;
    for(int &i=iter[u]; i<G[u].size(); i++)
    {
        edge &e=es[G[u][i]];
        if(dist[u]+1==dist[e.to]&&(d=DFS(e.to,t,min(f,e.cap-e.flow)))>0)
        {
            e.flow+=d;
            es[G[u][i]^1].flow-=d;
            flow+=d;
            f-=d;
            if (f==0) break;
        }
    }
    return flow;
}
int Maxflow(int s,int t)
{
    int flow=0;
    while(BFS(s,t))
    {
        memset(iter,0,sizeof(iter));
        int d=0;
        while(d=DFS(s,t,inf)) flow+=d;
    }
    return flow;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m,k;
        scanf("%d%d%d",&n,&m,&k);
        int s=0,t=n*m+1;
        for(int i=1; i<=n; i++)
        {
            addedge(s,(i-1)*m+1,inf);
            for(int j=1; j<=m; j++)
            {
                int x;
                scanf("%d",&x);
                if(j<m) addedge((i-1)*m+j,(i-1)*m+j+1,10000-x);
                else addedge((i-1)*m+j,t,10000-x);
            }
        }
        for(int i=1; i<=k; i++)
        {
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            for(int i=1; i<=m; i++)
            {
                for(int j=1; j<=m; j++)
                {
                    if(i-j>z)
                    {
                        if(j<m) addedge((x-1)*m+i,(y-1)*m+j+1,inf);
                        else addedge((x-1)*m+i,t,inf);
                    }
                }
            }
        }
        int ans=Maxflow(s,t);
        if(ans>=inf) printf("-1
");
        else printf("%d
",10000*n-ans);
        init(n*m);
    }
    return 0;
}