Luogu P2057 [SHOI2007]善意的投票

题目链接 \(Click\) \(Here\)

考虑模型转换。变成文理分科二选一带收益模型，就一波带走了。

如果没有见过这个模型的话，这里讲的很详细。

#include <bits/stdc++.h>
using namespace std;

#define LL long long
const int N = 400010;
const int M = 800010;
const int INF = 0x3f3f3f3f;
const LL INFF = 0x3f3f3f3f3f3f3f3f;

int n, m, cnt = -1, head[N];

struct edge {
	int nxt, to; LL f;
}e[M];

void add_len (int u, int v, LL f) {
	e[++cnt] = (edge) {head[u], v, f}; head[u] = cnt;
	e[++cnt] = (edge) {head[v], u, 0}; head[v] = cnt;
}

int node (int x) {return x;}
int f1 (int x) {return n + m * 0 + x;}
int f2 (int x) {return n + m * 1 + x;} 

queue <int> q;
int cur[N], deep[N];

bool bfs (int s, int t) {
	memcpy (cur, head, sizeof (head));
	memset (deep, 0x3f, sizeof (deep));
	deep[s] = 0; q.push (s);
	while (!q.empty ()) {
		int u = q.front (); q.pop ();
		for (int i = head[u]; ~i; i = e[i].nxt) {
			int v = e[i].to;
			if (deep[v] == INF && e[i].f) {
				deep[v] = deep[u] + 1;
				q.push (v);
			}
 		}
	}
	return deep[t] != INF;
}

LL dfs (int u, int t, LL lim) {
	if (u == t || !lim) {
		return lim;
	}
	int tmp = 0, flow = 0;
	for (int &i = cur[u]; ~i; i = e[i].nxt) {
		int v = e[i].to;
		if (deep[v] == deep[u] + 1) {
			tmp = dfs (v, t, min (lim, e[i].f));
			lim -= tmp;
			flow += tmp;
			e[i ^ 0].f -= tmp;
			e[i ^ 1].f += tmp;
			if (!lim) break;
		}
	}
	return flow;
}

int main () {
	memset (head, -1, sizeof (head));
	cin >> n >> m;
	int s = f2 (m + 1), t = f2 (m + 2); 
	for (int i = 1; i <= n; ++i) {
		static int cho;
		cin >> cho;
		if (cho == 0) {
			add_len (s, node (i), INF + 1);
			add_len (node (i), t, INF + 0);
		} else {
			add_len (s, node (i), INF + 0);
			add_len (node (i), t, INF + 1);
		}
		//s -> 0, t -> 1
	}
	for (int i = 1; i <= m; ++i) {
		static int x, y;
		cin >> x >> y;
		add_len (s, f1 (i), 1); add_len (f1 (i), node (x), INFF); add_len (f1 (i), node (y), INFF);
		add_len (f2 (i), t, 1); add_len (node (x), f2 (i), INFF); add_len (node (y), f2 (i), INFF);
	}
	LL min_cut = 0;
	while (bfs (s, t)) {
		min_cut += dfs (s, t, INFF);
	}
	LL ans = (min_cut - n * INF - m);
	cout << ans << endl;
}