题意：2个人从1走到n，假设一条路第一次走则是价值di，假设第二次还走这条路则须要价值di+ai，要你输出2个人到达终点的最小价值！

太水了！一条边建2次就OK了。第一次价值为di，第二次为ai+di，加入源点汇点跑最小费用最大流就OK了！

AC代码：

#include<iostream>

#include<string>

#include<algorithm>

#include<cstdlib>

#include<cstdio>

#include<set>

#include<map>

#include<vector>

#include<cstring>

#include<stack>

#include<cmath>

#include<queue>

using namespace std;

#define CL(x,v); memset(x,v,sizeof(x));

#define INF 0x3f3f3f3f

#define LL long long

#define REP(i,r,n) for(int i=r;i<=n;i++)

#define RREP(i,n,r) for(int i=n;i>=r;i--)

int sumFlow;

const int MAXN = 600+10;

const int MAXM = 4000+10;

struct Edge

{

    int u, v, cap, cost;

    int next;

}edge[MAXM<<2];

int NE;

int head[MAXN], dist[MAXN], pp[MAXN];

bool vis[MAXN];

void init()

{

    NE = 0;

    memset(head, -1, sizeof(head));

}

void addedge(int u, int v, int cap, int cost)

{

    edge[NE].u = u;

    edge[NE].v = v;

    edge[NE].cap = cap;

    edge[NE].cost = cost;

    edge[NE].next = head[u];

    head[u] = NE++;

    edge[NE].u = v;

    edge[NE].v = u;

    edge[NE].cap = 0;

    edge[NE].cost = -cost;

    edge[NE].next = head[v];

    head[v] = NE++;

}

bool SPFA(int s, int t, int n)

{

    int i, u, v;

    queue <int> qu;

    memset(vis,false,sizeof(vis));

    memset(pp,-1,sizeof(pp));

    for(i = 0; i <= n; i++) dist[i] = INF;

    vis[s] = true; dist[s] = 0;

    qu.push(s);

    while(!qu.empty())

    {

        u = qu.front(); qu.pop(); vis[u] = false;

        for(i = head[u]; i != -1; i = edge[i].next)

        {

            v = edge[i].v;

            if(edge[i].cap && dist[v] > dist[u] + edge[i].cost)

            {

                dist[v] = dist[u] + edge[i].cost;

                pp[v] = i;

                if(!vis[v])

                {

                    qu.push(v);

                    vis[v] = true;

                }

            }

        }

    }

    if(dist[t] == INF) return false;

    return true;

}

int MCMF(int s, int t, int n)

{

    int flow = 0;

    int i, minflow, mincost;

    mincost = 0;

    while(SPFA(s, t, n))

    {

        minflow = INF + 1;

        for(i = pp[t]; i != -1; i = pp[edge[i].u])

            if(edge[i].cap < minflow)

                minflow = edge[i].cap;

        flow += minflow;

        for(i = pp[t]; i != -1; i = pp[edge[i].u])

        {

            edge[i].cap -= minflow;

            edge[i^1].cap += minflow;

        }

        mincost += dist[t] * minflow;

    }

    sumFlow = flow;

    return mincost;

}

int main()

{

    int n,m;

    int cas=1;

    while(~scanf("%d%d",&n,&m))

    {

        init();

        int s=0;

        int t=n+1;

        for(int i=0;i<m;i++)

        {

            int u,v,w1,w2;

            scanf("%d %d %d %d",&u,&v,&w1,&w2);

            addedge(u,v,1,w1);

            addedge(u,v,1,w1+w2);

        }

        addedge(s,s+1,2,0);

        addedge(t-1,t,2,0);

        int ans=MCMF(s,t,t);

        printf("Case %d: %d\n",cas++,ans);

    }

    return 0;

}