計蒜客習題-班長競選 最小割 “二選一 方案不同有額外開銷”模型

設源點S，匯點T

對於所有贊成的人 從S連一條邊到他們對應的點上 容量設爲1

對於所有反對的人 從他們對應的點上連一條邊到T 容量爲1

對於所有有朋友 關係的a和b 在他們之間連一條無向邊 容量爲1

最小割可以用dinic跑出來 即所求答案
只要記好朋友之間的衝突總數就好了

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const int maxn = 304;
const int inf = 0x7f7f7f7f;
const int maxm = 4500;
int n, m,S,T;
struct edge {
	int v, c, next;
}e[maxn*maxn<<1];
int p[maxn], eid = 0,d[maxn];
inline void ins(int u, int v, int c) {
	e[eid].v = v; e[eid].c = c; e[eid].next = p[u]; p[u] = eid++;
}
inline void addedge(int u, int v, int c) {
	ins(u, v, c);
	ins(v, u, 0);
}
bool bfs() {
	memset(d, -1, sizeof(d));
	queue<int> q;
	q.push(S);
	d[S] = 0;
	while (!q.empty()) {
		int u = q.front(); q.pop();
		for (int i = p[u]; ~i; i = e[i].next) {
			int v = e[i].v;
			if (e[i].c > 0 && d[v] == -1) {
				d[v] = d[u] + 1;
				q.push(v);
			}
		}
	}
	return (d[T] != -1);
}
int dfs(int u, int flow) {
	if (u == T) return flow;
	int res = 0;
	for (int i = p[u]; ~i; i = e[i].next) {
		int v = e[i].v;
		if (e[i].c > 0 && d[v] == d[u] + 1) {
			int tmp = dfs(v, min(flow, e[i].c));
			res += tmp;
			flow -= tmp;
			e[i].c -= tmp;
			e[i ^ 1].c += tmp;
			if (flow == 0) break;	//飽和
		}
	}
	if (res == 0) d[u] = -1;
	return res;
}
int dinic() {
	int res = 0;
	while (bfs()) {
		res += dfs(S, 0x7f7f7f7f);
		//printf("%d\n", res);
	}
	return res;
}
inline void init() {
	memset(p, -1, sizeof(p));
	eid = 0;
}
inline int read() {
	int s = 0, f = 1; char c = getchar(); while (c<'0' || c>'9') { if (c == '-') f = -1; c = getchar(); }
	while (c >= '0'&&c <= '9') { s = s * 10 + c - '0'; c = getchar(); }
	return s*f;
}
int main() {
	n = read(); m = read();
	int a, b;
	S = 0; T = n + 1;
	init();

	for (int i = 1; i <= n; i++) {
		if (read() == 1) addedge(S, i, 1);
		else addedge(i, T, 1);
	}
	for (int i = 1; i <= m; i++) {
		a = read(); b = read();
		addedge(a, b, 1);		//好朋友意願不同也要計算衝突
		addedge(b, a, 1);
	}
	printf("%d", dinic());
	//getchar();
	return 0;
}