版权声明：转载请注明出处。 https://blog.csdn.net/u014427196/article/details/48281573

最大权闭合图学习资料： [

http://www.cnblogs.com/wuyiqi/archive/2012/03/12/2391960.html
](http://www.cnblogs.com/wuyiqi/archive/2012/03/12/2391960.html)

最大权 = 正的权值的和-建图后的最小割的容量。

题目连接： http://acm.hdu.edu.cn/showproblem.php?pid=3879

题意：有n个点，m个选择，建造n个点各自需要一定花费，每个选择有一定的获利，会选择两个点。求最大的获利

代码：

#include <iostream>  
#include <algorithm>  
#include <set>  
#include <map>  
#include <string.h>  
#include <queue>  
#include <sstream>  
#include <stdio.h>  
#include <math.h>  
#include <stdlib.h>  
#include <string>

using namespace std;

const int MAXN = 101000;//点数的最大值
const int MAXM = 4001000;//边数的最大值
const int INF = 0x3f3f3f3f;

struct Edge
{
    int to, next, cap, flow;
}edge[MAXM];//注意是MAXM

int tol;
int head[MAXN];
int gap[MAXN], dep[MAXN], pre[MAXN], cur[MAXN];

void init()
{
    tol = 0;
    memset(head, -1, sizeof(head));
}
//加边，单向图三个参数，双向图四个参数
void addedge(int u, int v, int w, int rw = 0)
{
    edge[tol].to = v; edge[tol].cap = w; edge[tol].next = head[u];
    edge[tol].flow = 0; head[u] = tol++;
    edge[tol].to = u; edge[tol].cap = rw; edge[tol].next = head[v];
    edge[tol].flow = 0; head[v] = tol++;
}
//输入参数：起点、终点、点的总数
//点的编号没有影响，只要输入点的总数
int sap(int start, int end, int N)
{
    memset(gap, 0, sizeof(gap));
    memset(dep, 0, sizeof(dep));
    memcpy(cur, head, sizeof(head));
    int u = start;
    pre[u] = -1;
    gap[0] = N;
    int ans = 0;
    while (dep[start] < N)
    {
        if (u == end)
        {
            int Min = INF;
            for (int i = pre[u]; i != -1; i = pre[edge[i ^ 1].to])
                if (Min > edge[i].cap - edge[i].flow)
                    Min = edge[i].cap - edge[i].flow;
            for (int i = pre[u]; i != -1; i = pre[edge[i ^ 1].to])
            {
                edge[i].flow += Min;
                edge[i ^ 1].flow -= Min;
            }
            u = start;
            ans += Min;
            continue;
        }
        bool flag = false;
        int v;
        for (int i = cur[u]; i != -1; i = edge[i].next)
        {
            v = edge[i].to;
            if (edge[i].cap - edge[i].flow && dep[v] + 1 == dep[u])
            {
                flag = true;
                cur[u] = pre[v] = i;
                break;
            }
        }
        if (flag)
        {
            u = v;
            continue;
        }
        int Min = N;
        for (int i = head[u]; i != -1; i = edge[i].next)
            if (edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
            {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        gap[dep[u]]--;
        if (!gap[dep[u]])return ans;
        dep[u] = Min + 1;
        gap[dep[u]]++;
        if (u != start) u = edge[pre[u] ^ 1].to;
    }
    return ans;
}

int n, m;
const int inf = 1e9;

int main()
{
    while (scanf("%d%d", &n, &m) != EOF)
    {
        init();
        int x;
        int s = 0, t = n + m + 1;
        for (int i = 1;i <= n;i++)
        {
            scanf("%d", &x);
            addedge(i, t, x);
        }
        int a, b, c;
        int tmp = 0;
        for (int i = 1;i <= m;i++)
        {
            scanf("%d%d%d", &a, &b, &c);
            tmp += c;
            addedge(s, i + n, c);
            addedge(i + n, a, inf);
            addedge(i + n, b, inf);
        }
        int ans = tmp - sap(s, t, n + m + 2);
        printf("%d\n",ans);
    }
    return 0;
}