ZrOJ #878. 小K与赞助 (网络流)

傻逼最大费用流：

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两棵树分别流，最后汇合。

CODE

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
const int MAXP = 1510;
const int MAXN = 505;
const int MAXM = 5005;
const int INF = 0x3f3f3f3f;
int n, a[MAXN], rt[2], q[2], fa[2][MAXN], lim[2][MAXN];
vector<int>G[2][MAXN];

int S, T, ans;

void dfs(int o, int u) {
	int siz = G[o][u].size(), v;
	for(int i = 0; i < siz; ++i)
		if((v=G[o][u][i]) != fa[o][u]) {
			fa[o][v] = u;
			dfs(o, v);
		}
}


int info[MAXP], fir[MAXP], to[MAXM], nxt[MAXM], c[MAXM], w[MAXM], cnt=1;
inline void link(int u, int v, int cc, int ww) {
	to[++cnt] = v, nxt[cnt] = fir[u], fir[u] = cnt; c[cnt] = cc, w[cnt] = ww;
	to[++cnt] = u, nxt[cnt] = fir[v], fir[v] = cnt; c[cnt] = 0, w[cnt] = -ww;
}
int dis[MAXP];
bool inq[MAXP];
inline bool spfa() {
	static queue<int> q;
	memset(dis, -1, sizeof dis);
	dis[S] = 0; q.push(S);
	while(!q.empty()) {
		int u = q.front(); q.pop(); inq[u] = 0;
		for(int i = fir[u], v; i; i = nxt[i])
			if(c[i] && dis[v=to[i]] < dis[u] + w[i]) {
				dis[v] = dis[u] + w[i];
				if(!inq[v]) inq[v] = 1, q.push(v);
			}
	}
	return ~dis[T];
}

bool vis[MAXP];

int Aug(int u, int Max) {
	if(u == T) { ans += Max * dis[T]; return Max; }
	vis[u] = 1;
	int delta, flow = 0;
	for(int &i = info[u], v; i; i = nxt[i])
		if(c[i] && dis[v=to[i]] == dis[u] + w[i] && !vis[v] && (delta=Aug(v, min(Max-flow, c[i])))) {
			flow += delta, c[i] -= delta, c[i^1] += delta;
			if(flow == Max) break;
		}
	vis[u] = 0; return flow;
}

inline void Max_Cost_flow() {
	while(spfa())
		memcpy(info, fir, sizeof fir), Aug(S, INF);
}

int main () {
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i)
		scanf("%d", &a[i]);
	for(int o = 0; o < 2; ++o) {
		scanf("%d", &rt[o]);
		for(int i = 1, x, y; i < n; ++i)
			scanf("%d%d", &x, &y), G[o][x].pb(y), G[o][y].pb(x);
		dfs(o, rt[o]);
	}
	memset(lim, 0x3f, sizeof lim);
	for(int o = 0; o < 2; ++o) {
		scanf("%d", &q[o]);
		for(int i = 1, x, y; i <= q[o]; ++i)
			scanf("%d%d", &x, &y), lim[o][x] = y;
	}
	S = 0, T = 3*n + 1;
	for(int i = 1; i <= n; ++i) {
		link(fa[0][i], i, lim[0][i], 0);
		link(fa[1][i]?n+fa[1][i]:0, n+i, lim[1][i], 0);
		link(i, 2*n+i, 1, 0);
		link(n+i, 2*n+i, 1, 0);
		link(2*n+i, T, 1, a[i]);
	}
	Max_Cost_flow();
	printf("%d
", ans);
}