POJ 2135 Farm Tour 最小费用流

两条路不能有重边，既每条边的容量是1。求流量为2的最小费用即可。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c)
{
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c)
{
    return max(max(a,b),max(a,c));
}
void debug()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in","r",stdin);
    // freopen("d:\\out1.txt","w",stdout);
#endif
}
int getch()
{
    int ch;
    while((ch=getchar())!=EOF)
    {
        if(ch!=' '&&ch!='\n')return ch;
    }
    return EOF;
}

struct Edge
{
    int from,to,cost,cap;
};
const int maxn = 3111;

vector<int> g[maxn];
vector<Edge> edge;
int n,m,s,t;
void init()
{
    for(int i = 1; i <= n; i++)
        g[i].clear();
    edge.clear();
}
void add(int from, int to, int cost, int cap)
{
    edge.push_back((Edge){from, to, cost, cap});
    g[from].push_back(edge.size() - 1);
    edge.push_back((Edge){to, from, -cost, 0});
    g[to].push_back(edge.size() - 1);
}

int d[maxn];
int inq[maxn];
int road[maxn];

int SPFA()
{
    queue<int> q;
    q.push(s);
    memset(d, INF, sizeof(d));
    memset(inq, 0, sizeof(inq));
    inq[s] = true;
    d[s] = 0;
    road[s] = -1;
    while(!q.empty())
    {
        int x = q.front(); q.pop();
        inq[x] = false;
        for(int i = 0; i < g[x].size(); i++)
        {
            Edge &e = edge[g[x][i]];
            if(e.cap>0 && d[x] + e.cost < d[e.to])
            {
                d[e.to] = d[x] + e.cost;
                road[e.to] = g[x][i];

                if(!inq[e.to])
                {
                    inq[e.to] = true;
                    q.push(e.to);
                }
            }
        }
    }
    return d[t];
}
int max_cost_flow()
{
    int flow = 2;
    int cost = 0;
    while(flow)
    {
        int d = SPFA();
        int f = flow;
        for(int e = road[t]; e != -1; e = road[edge[e].from])
        {
            Edge &E = edge[e];
            f = min(f, E.cap);
        }
        flow -= f;
        cost += d * f;
        for(int e = road[t]; e != -1; e = road[edge[e].from])
        {
            edge[e].cap -= f;
            edge[e^1].cap += f;
        }
    }
    return cost;
}
int main()
{
    debug();
    while(scanf("%d%d", &n, &m) != EOF)
    {
        init();
        for(int i = 1; i <= m; i++)
        {
            int from,to,cost;
            scanf("%d%d%d", &from, &to, &cost);
            add(from, to, cost, 1);
            add(to, from, cost, 1);
        }
        s=1;
        t=n;
        printf("%d\n", max_cost_flow());
    }
    return 0;
}