洛谷 P4307 [JSOI2009]球队收益 / 球队预算（最小费用最大流）

题面

luogu

题解

最小费用最大流

先假设剩下(m)场比赛，双方全输。
考虑(i)赢一局的贡献
(C_i*(a_i+1)^2+D_i*(b_i-1)^2-C_i*a_i^2-D_i*b_i^2)
(=C _i+2*a_i*C_i+D_i-2*b_i*D_i)

建(m)个点限制每场比赛只有一个人赢，自(S)连一条((1, 0))的边

然后从这(m)的点连向和比赛有关的两个点一条((1, 0))的边

考虑关于(t)的边

因为(a_i)和(b_i)会变，不能直接连一条边

然后我们观察一下上面那个式子，
显然，赢得更多，贡献就增长越大

那么，拆边，我们可以对于每个点连向(t)很多条边，容量为(1)，价值是赢了对应场数的贡献

贪心地想，因为赢得更多，贡献增长越大，要最小费用，我们会先走赢得少的边
这样累加答案刚好满足条件呢

Code

#include<bits/stdc++.h>

#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 = 10000, inf = 2147483647;

struct node {
	int to, nxt, w, v;
}g[200*N];
int last[N], gl = 1;
void add(int x, int y, int w, int v) {
	g[++gl] = (node) {y, last[x], w, v};
	last[x] = gl;
	g[++gl] = (node) {x, last[y], 0, -v};
	last[y] = gl;
}

queue<int> q;
int pre[N], dis[N], from[N], s, t;
bool vis[N];

bool spfa() {
	memset(dis, 127, sizeof(dis));
	dis[s] = 0;
	q.push(s);
	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 && dis[v] > dis[u] + g[i].v) {
				dis[v] = dis[u] + g[i].v; pre[v] = u; from[v] = i;
				if (!vis[v]) {
					vis[v] = 1;
					q.push(v);
				}
			}
		}
		vis[u] = 0;
	}
	return dis[t] != dis[0];
}

int Mcmf() {
	int ans = 0;
	while (spfa()) {
		int di = inf;
		for (int i = t; i != s; i = pre[i]) di = min(di, g[from[i]].w);
		ans += di*dis[t];
		for (int i = t; i != s; i = pre[i]) g[from[i]].w -= di, g[from[i]^1].w += di;
	}
	return ans;
}
int a[N], b[N], c[N], d[N], cnt[N];

int main() {
	int n, m, sum = 0;
	read(n), read(m);
	s = m+n+1, t = s+1;
	for (int i = 1; i <= n; i++)
		read(a[i]), read(b[i]), read(c[i]), read(d[i]);
	for (int i = 1; i <= m; i++) {
		int x, y;
		read(x), read(y);
		b[x]++, b[y]++;
		cnt[x]++; cnt[y]++;
		add(s, i, 1, 0);
		add(i, x+m, 1, 0);
		add(i, y+m, 1, 0);
	}
	for (int i = 1; i <= n; i++)
		sum += c[i]*a[i]*a[i] + d[i]*b[i]*b[i];
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= cnt[i]; j++) {
			add(i+m, t, 1, c[i]+2*a[i]*c[i]+d[i]-2*b[i]*d[i]);
			b[i]--; a[i]++;
		}
	}
	printf("%d
", sum+Mcmf());
	return 0;
}