POJ1149 PIGS

题目链接

题解

对于一个人有(k)个钥匙,

然后这(k)个猪圈就等于合并了

那么我们考虑转换问题,

一个人先把(k)个猪圈的猪全买了,再转让给后面的人

具体连边看代码..

Code

#include<cstdio>
#include<queue>
#include<cstring>

#define LL long long
#define RG register

using namespace std;
template<class T> inline void read(T &x) {
	x = 0; RG char c = getchar(); bool f = 0;
	while (c != '-' && (c < '0' || c > '9')) c = getchar(); if (c == '-') c = getchar(), f = 1;
	while (c >= '0' && c <= '9') x = x*10+c-48, c = getchar();
	x = f ? -x : x;
	return ;
}
template<class T> inline void write(T x) {
	if (!x) {putchar(48);return ;}
	if (x < 0) x = -x, putchar('-');
	int len = -1, z[20]; while (x > 0) z[++len] = x%10, x /= 10;
	for (RG int i = len; i >= 0; i--) putchar(z[i]+48);return ;
}
const int N = 2010, inf = 2147483647;
int s, t;

struct node {
	int to, nxt, w;
}g[2000010];
int last[N], gl = 1, cur[N];
void add(int x, int y, int z) {
	g[++gl] = (node) {y, last[x], z};
	last[x] = gl;
	g[++gl] = (node) {x, last[y], 0};
	last[y] = gl;
}

queue<int> q;

int dep[N];

bool bfs() {
	q.push(s);
	memset(dep, 0, sizeof(dep));
	dep[s] = 1;
	while (!q.empty()) {
		int u = q.front(); q.pop();
		for (int i = last[u]; i; i = g[i].nxt) {
			int v = g[i].to;
			if (g[i].w > 0 && !dep[v])
				dep[v] = dep[u] + 1, q.push(v);
		}
	}
	return dep[t];
}

int dfs(int u, int d) {
	if (u == t) return d;
	for (int &i = cur[u]; i; i = g[i].nxt) {
		int v = g[i].to;
		if (dep[v] == dep[u]+1 && g[i].w) {
			int di = dfs(v, min(d, g[i].w));
			if (di) {
				g[i].w -= di;
				g[i ^ 1].w += di;
				return di;
			}
		}
	}
	return 0;
}

int dinic() {
	int ans = 0;
	while (bfs()) {
		for (int i = 1; i <= t; i++)
			cur[i] = last[i];
		while (int d = dfs(s, inf))
			ans += d;
	}
	return ans;
}

int p[N], id[N], a[N][N], b[N];

int main() {
	int m, n;
	read(n), read(m);
	s = n + 1; t = s + 1;
	for (int i = 1; i <= n; i++) read(p[i]);
	for (int i = 1; i <= m; i++) {
		read(a[i][0]);
		for (int j = 1; j <= a[i][0]; j++) read(a[i][j]);
		read(b[i]);
	}
	for (int i = 1; i <= m; i++) {
		add(i, t, b[i]);
		for (int j = 1; j <= a[i][0]; j++) {
			if (!id[a[i][j]]) add(s, i, p[a[i][j]]);
			else add(id[a[i][j]], i, inf);
			id[a[i][j]] = i;
		}
	}
	printf("%d
", dinic());
	return 0;
}