Mafia

戳戳看题
这道题的核心就是教你建双向边。我们发现本题的权值是在点上的。那么我们可以将1个点拆成2个点（一个点是入，一个点是出），中间的权值为val[i]。

上马！

#include<cstdio>
#include<cctype>
#include<queue>
#include<iostream>
#include<cstring>
using namespace std;

const int N = 90002, lim = (1 << 30) - 1;
int val[N], s, t, d[N], cur[N], n, m, cnt, nxt[N], flow[N], to[N], head[N];

inline int read() {
    int x = 0, f = 1;
    char s = getchar();
    while(! isdigit(s)) {
        if(s == '-')
            f = -1;
        s = getchar();
    }
    while(isdigit(s)) {
        x = (x << 1) + (x << 3) + (s ^ 48);
        s = getchar();
    }
    return x * f;
}

inline void print(int x) {
    if(x < 0) {
        putchar('-');
        x = -x;
    }
    if(x > 9)
        print(x / 10);
    putchar(x % 10 + '0');
}

void addEdge(const int u, const int v, const int w) {
	to[++ cnt] = v;
	nxt[cnt] = head[u];
	flow[cnt] = w;
	head[u] = cnt;
}

int dfs(const int u, const int Flow) {
	if(u == t)
		return Flow;
	for(int &i = cur[u]; ~i; i = nxt[i]) {
		int v = to[i];
		if(d[v] == d[u] + 1 && flow[i] > 0) {
			int dis = dfs(v, min(Flow, flow[i]));
			if(dis > 0) {
				flow[i] -= dis;
				flow[i ^ 1] += dis;
				return dis;
			}
		}
	}
	return 0;
}

bool bfs() {
	queue <int> q;
	while(! q.empty())
		q.pop();
	memset(d, 0, sizeof d);
	d[s] = 1;
	q.push(s);
	while(! q.empty()) {
		int u = q.front();
		q.pop();
		for(int i = head[u]; ~i; i = nxt[i]) {
			int v = to[i];
			if(d[v] == 0 && flow[i] > 0) {
				d[v] = d[u] + 1;
				q.push(v);
			}
		}
	}
	return d[t] > 0;
}

int dinic() {
	int ans = 0, dis;
	while(bfs()) {
		for(int i = 1; i <= (n << 1); ++ i)
			cur[i] = head[i];
		while(dis = dfs(s, lim))
			ans += dis;
	}
	return ans;
}

int main() {
	int a, b;
	while(scanf("%d", &n) != EOF) {
		m = read();
		s = read();
		t = read() + n;
		cnt = -1;
		memset(head, -1, sizeof head);
		for(int i = 1; i <= n; ++ i) {
			val[i] = read();
			addEdge(i, i + n, val[i]);
			addEdge(i + n, i, 0);
		}
		for(int i = 1; i <= m; ++ i) {
			a = read();
			b = read();
			addEdge(b + n, a, lim);
			addEdge(a, b + n, 0);
			addEdge(a + n, b, lim);
			addEdge(b, a + n, 0);
		}
		printf("%d\n", dinic());
	}
	return 0;
}

来源：CSDN

作者：我只是想康康昵称会显示在哪里

链接：https://blog.csdn.net/LBCLZMB/article/details/103481249