【网络流】 TOJ 3854. Haitang2

先对0号节点做强连通，那么不在0号节点的强连通分量的点没有意义，可以舍弃掉。在0号节点的强连通分量里面的所有边都是至少走一次的，那么我们先统计答案，可以看出需要求的花的答案已经求出来了。原来强连通分量里面的边保留不变，流量设为无穷，费用为该边的费用，然后对连通分量所有点按度差建边，使全图入度和初度相同。。。跑出来的费用加上之前的费用就是最终答案。。。

#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 305
#define maxm 20005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL; 
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

struct Edge
{
	int v, c, w, next;
	Edge(int v = 0, int c = 0, int w = 0, int next = 0) : v(v), c(c), w(w), next(next) {}
}E[maxm];

struct edge
{
	int u, v, a, b, next;
	edge(int u = 0, int v = 0, int a = 0, int b = 0, int next = 0) : u(u), v(v), a(a), b(b), next(next) {}
}EE[maxm];

int in[maxn];
int out[maxn];
int dfn[maxn];
int low[maxn];
stack<int> ss;
queue<int> q;
int H[maxn], cntE;
int h[maxn], cntEE;
int cap[maxn];
int dis[maxn];
int cur[maxn];
int vis[maxn];
int ins[maxn];
int scc[maxn];
int dfs_clock, n, m;
int flow, cost, s, t, T;

void add_edges(int u, int v, int a, int b)
{
	EE[cntEE] = edge(u, v, a, b, h[u]);
	h[u] = cntEE++;
}

void addedges(int u, int v, int c, int w)
{
	E[cntE] = Edge(v, c, w, H[u]);
	H[u] = cntE++;
	E[cntE] = Edge(u, 0, -w, H[v]);
	H[v] = cntE++;
}

void init()
{
	cntEE = cntE = T = dfs_clock = 0;
	memset(h, -1, sizeof h);
	memset(H, -1, sizeof H);
	memset(in, 0, sizeof in);
	memset(out, 0, sizeof out);
	memset(scc, 0, sizeof scc);
	memset(ins, 0, sizeof ins);
	memset(vis, 0, sizeof vis);
	memset(dfn, 0, sizeof dfn);
}

bool spfa()
{
	memset(dis, INF, sizeof dis);
	q.push(s);
	vis[s] = ++T;
	cap[s] = INF;
	dis[s] = 0;
	cur[s] = -1;
	while(!q.empty()) {
		int u = q.front();
		q.pop();
		vis[u] = T - 1;
		for(int e = H[u]; ~e; e = E[e].next) {
			int v = E[e].v, c = E[e].c, w = E[e].w;
			if(c && dis[u] + w < dis[v]) {
				dis[v] = dis[u] + w;
				cap[v] = min(c, cap[u]);
				cur[v] = e;
				if(vis[v] != T) {
					vis[v] = T;
					q.push(v);
				}
			}
		}
	}
	if(dis[t] == INF) return false;
	flow += cap[t];
	cost += cap[t] * dis[t];
	for(int e = cur[t]; ~e; e = cur[E[e ^ 1].v]) {
		E[e].c -= cap[t];
		E[e ^ 1].c += cap[t];
	}
	return true;
}

int mcmf()
{
	flow = cost = 0;
	while(spfa());
	return cost;
}

void tarjan(int u)
{
	dfn[u] = low[u] = ++dfs_clock;
	ss.push(u), ins[u] = true;
	for(int e = h[u]; ~e; e = EE[e].next) {
		int v = EE[e].v;
		if(!dfn[v]) {
			tarjan(v);
			low[u] = min(low[u], low[v]);
		}
		else if(ins[v]) low[u] = min(low[u], dfn[v]);
	}
	if(low[u] == dfn[u]) {
		int t;
		do {
			t = ss.top();
			ss.pop();
			ins[t] = false;
			if(u == 0) scc[t] = true;
		}while(t != u);
	}
}

void work()
{
	scanf("%d%d", &n, &m);
	
	for(int i = 1; i <= m; i++) {
		int u, v, a, b;
		scanf("%d%d%d%d", &u, &v, &a, &b);
		add_edges(u, v, a, b);
	}

	tarjan(0);

	int res1 = 0, res2 = 0;
	for(int i = 0; i < cntEE; i++) {
		int u = EE[i].u, v = EE[i].v, a = EE[i].a, b = EE[i].b;
		if(scc[u] && scc[v]) {
			in[v]++, out[u]++;
			res1 += a, res2 += b;
			addedges(u, v, INF, a);
		}
	}

	s = n + 2, t = n+1;
	for(int i = 0; i < n; i++) if(scc[i]) {
		if(out[i] > in[i]) addedges(i, t, out[i] - in[i], 0);
		else addedges(s, i, in[i] - out[i], 0);
	}
	
	res1 += mcmf();
	printf("%d %d\n", res2, res1);
}

int main()
{
	int _;
	scanf("%d", &_);
	while(_--) {
		init();
		work();
	}
	
	return 0;
}