【BZOJ1449】[JSOI2009]球队收益【最小费用最大流】【单调增函数建图】

【题目链接】

【POPOQQQ的题解】%一发建图姿势。

/* Pigonometry */
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

const int maxn = 10005, maxm = 500005, maxq = 100000, inf = 0x3f3f3f3f;

int n, m, head[maxn], cnt, depth[maxn], way[maxn], bg, ed, q[maxq], C[maxn], D[maxn], win[maxn], lose[maxn], du[maxn];
bool vis[maxn];

struct _edge {
	int v, w, c, next;
} g[maxm << 1];

inline int iread() {
	int f = 1, x = 0; char ch = getchar();
	for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
	for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	return f * x;
}

inline void add(int u, int v, int w, int c) {
	g[cnt] = (_edge){v, w, c, head[u]};
	head[u] = cnt++;
}

inline void insert(int u, int v, int w, int c) {
	add(u, v, w, c); add(v, u, 0, -c);
}

inline bool spfa() {
	for(int i = 0; i <= ed; i++) depth[i] = inf;
	int h = 0, t = 0, u, i; depth[q[t++] = bg] = 0;
	while(h != t) for(i = head[u = q[h++]], vis[u] = 0; ~i; i = g[i].next) if(g[i].w && depth[g[i].v] > depth[u] + g[i].c) {
		depth[g[i].v] = depth[u] + g[i].c;
		way[g[i].v] = i;
		if(!vis[g[i].v]) vis[q[t++] = g[i].v] = 1;
	}
	return depth[ed] != inf;
}

inline int back() {
	int flow = inf, res = 0;
	for(int u = ed; u != bg; u = g[way[u] ^ 1].v) flow = min(flow, g[way[u]].w);
	for(int u = ed; u != bg; u = g[way[u] ^ 1].v) g[way[u]].w -= flow, g[way[u] ^ 1].w += flow, res += g[way[u]].c * flow;
	return res;
}

int main() {
	n = iread(); m = iread(); bg = 0; ed = n + m + 1;
	for(int i = 0; i <= ed; i++) head[i] = -1; cnt = 0;

	for(int i = 1; i <= n; i++) {
		win[i] = iread(); lose[i] = iread();
		C[i] = iread(); D[i] = iread();
	}
	for(int i = 1; i <= m; i++) {
		int u = iread(), v = iread();
		insert(u, n + i, 1, 0);
		insert(v, n + i, 1, 0);
		insert(n + i, ed, 1, 0);
		du[u]++; du[v]++;
	}

	LL ans = 0;
	for(int i = 1; i <= n; i++) {
		ans += (LL)C[i] * win[i] * win[i] + (LL)D[i] * (du[i] + lose[i]) * (du[i] + lose[i]);
		for(int j = 1; j <= du[i]; j++)
			insert(bg, i, 1, C[i] * (win[i] * 2 + j * 2 - 1) - D[i] * ((du[i] + lose[i]) * 2 - j * 2 + 1));
	}

	while(spfa()) ans += back();
	printf("%lld\n", ans);
	return 0;
}