【网络流】 POJ 2135 Farm Tour

费用流模板题。。。源点向出发点连一条容量为2,费用为0的边，终点向汇点连一条容量为1,费用为0的边。。。中间的无向边就连容量为1,费用为边的权值的边。。。跑费用流就行啦。。。

#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 1005
#define maxm 40005
#define eps 1e-10
#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
#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;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}
// head

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

queue<int> q;
int H[maxn], cntE;
int vis[maxn];
int dis[maxn];
int cap[maxn];
int cur[maxn];
int flow, cost, T, s, t;
int n, m;

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++;
}

bool spfa(void)
{
	memset(dis, INF, sizeof dis);
	cur[s] = -1;
	vis[s] = ++T;
	cap[s] = INF;
	dis[s] = 0;
	q.push(s);
	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[v] > dis[u] + w) {
				dis[v] = dis[u] + w;
				cap[v] = min(cap[u], c);
				cur[v] = e;
				if(vis[v] != T) {
					vis[v] = T;
					q.push(v);
				}
			}
		}
	}
	if(dis[t] == INF) return false;
	cost += cap[t] * dis[t];
	flow += cap[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(void)
{
	flow = cost = 0;
	while(spfa());
	return cost;
}

void init(void)
{
	T = 0;
	cntE = 0;
	memset(H, -1, sizeof H);
	memset(vis, 0, sizeof vis);
}

void read(void)
{
	int a, b, c;
	while(m--) {
		scanf("%d%d%d", &a, &b, &c);
		addedges(a, b, 1, c);
		addedges(b, a, 1, c);
	}
}

void work(void)
{
	s = 0, t = n + 1;
	addedges(s, 1, 2, 0);
	addedges(n, t, 2, 0);
	int ans = mcmf();
	printf("%d\n", ans);
}

int main(void)
{
	while(scanf("%d%d", &n, &m)!=EOF) {
		init();
		read();
		work();
	}

	return 0;
}