CodeForces-1082G Increasing Frequency

题目链接：CodeForces-1082G Increasing Frequency

题意

有$n$个点，$m$条边，每个点和每条边都有一个权值，要求你从所给的图中选出一个子图，使得$sum{val_{edges}}-sum{val_{vertices}}$的值最大，子图的边集需满足：每条边关联的两个结点都在子图的点集中。

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思路

闭合子图是选了结点 $u$ ，那么 $u$ 的所有后继结点也要选，而这道题是选了一条边，那么这条边的两个端点也要选，根据这个性质，把原图的边转换为结点，边权转换为点权，向两个端点连有向边，这样两个端点成为代表边的结点的后继结点，题目也就转换为了闭合子图问题。原图结点点权对答案贡献为负，所以点权取相反数改为负权值，边权对答案贡献为正，所以代表边的结点仍为正权值，题目即为求以上构建出的这个图的最大权闭合子图。

最大权闭合子图详解：传送门

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代码实现

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using std::queue;
typedef long long LL;
const int INF = 0x3f3f3f3f, N = 5010, M = 50010;
int head[N], d[N];
int n, m, s, t, tot;
LL maxflow;
struct Edge
{
    int to, cap, nex;
} edge[M];
queue<int> q;
void add(int x, int y, int z) {
    edge[++tot].to = y, edge[tot].cap = z, edge[tot].nex = head[x], head[x] = tot;
    edge[++tot].to = x, edge[tot].cap = 0, edge[tot].nex = head[y], head[y] = tot;
}
bool bfs() { // 在残存网络上构造分层图
    memset(d, 0, sizeof(d));
    while (q.size()) q.pop();
    q.push(s); d[s] = 1;
    while (q.size()) {
        int x = q.front(); q.pop();
        for (int i = head[x]; i; i = edge[i].nex) {
            int v = edge[i].to;
            if (edge[i].cap && !d[v]) {
                q.push(v);
                d[v] = d[x] + 1;
                if (v == t) return true;
            }
        }
    }
    return false;
}
int dinic(int x, int flow) {  // 在当前分层图上增广
    if (x == t) return flow;
    int rest = flow, k;
    for (int i = head[x]; i && rest; i = edge[i].nex) {
        int v = edge[i].to;
        if (edge[i].cap && d[v] == d[x] + 1) {
            k = dinic(v, std::min(rest, edge[i].cap));
            if (!k) d[v] = 0;  // 剪枝，去掉增广完毕的点(当前弧优化)
            edge[i].cap -= k;
            edge[i^1].cap += k;
            rest -= k;
        }
    }
    return flow - rest;
}

int main() {
    while (~scanf("%d %d", &n, &m)) {
        s = 0, t = n + m + 1;
        tot = 1, maxflow = 0;
        memset(head, 0, sizeof(head));
        for (int i = 1, val; i <= n; i++) {
            scanf("%d", &val);
            add(i, n + m + 1, val);
        }
        LL sum = 0;
        for (int i = 1, u, v, val; i <= m; i++) {
            scanf("%d %d %d", &u, &v, &val);
            sum += val;
            add(0, n + i, val);
            add(n + i, u, INF);
            add(n + i, v, INF);
        }
        while (bfs()) maxflow += dinic(s, INF);
        printf("%I64d
", sum - maxflow);
    }
    return 0;
}