csu 1506 Double Shortest Paths 最小费用最大流

题意：

两个人从1走到n

每条边走第一次花费是w1 第二次是w1+w2

问最少费用

思路：

印象中能纠正的就是最小费用最大流 就试着用了一下

每条遍分解成第一次走和第二次走两种情况

然后手动添加源点汇点 直接跑一遍模板就行了

（感谢kuangbin~

注意边都加了两遍 所以边的集合要开两倍

#include<bits/stdc++.h>
#define cl(a,b) memset(a,b,sizeof(a))
#define debug(x) cerr<<#x<<"=="<<(x)<<endl
using namespace std;

const int maxn = 500+10;
const int maxm = 4000+10;
const int inf = 0x3f3f3f3f;

struct Edge
{
    int to, next, cap, flow, cost;
}edge[maxm];

int n,m;
int head[maxn],tol;
int pre[maxn],dis[maxn];
bool vis[maxn];

void init()
{
    tol = 0;
    cl(head,-1);
}

void addedge(int u, int v, int cap, int cost)
{
    edge[tol].to=v;
    edge[tol].cap=cap;
    edge[tol].cost=cost;
    edge[tol].flow=0;
    edge[tol].next=head[u];
    head[u]=tol++;
    edge[tol].to=u;
    edge[tol].cap=0;
    edge[tol].cost=-cost;
    edge[tol].flow=0;
    edge[tol].next=head[v];
    head[v]=tol++;
}

bool spfa(int s, int t)
{
    queue<int>q;
    while(!q.empty()) q.pop();
    cl(dis,inf),cl(vis,false),cl(pre,-1);
    dis[s] = 0;
    vis[s] = true;
    q.push(s);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u] = false;
        for(int i = head[u]; i != -1; i = edge[i]. next)
        {
            int v = edge[i]. to;
            if(edge[i]. cap > edge[i]. flow &&
               dis[v] > dis[u] + edge[i]. cost )
            {
                dis[v] = dis[u] + edge[i]. cost;
                pre[v] = i;
                if(!vis[v])
                {
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    if(pre[t] == -1) return false;
    else return true;
}

void solve(int s, int t, int &cost)
{
    int flow = 0;
    cost = 0;
    while(spfa(s,t))
    {
        int Min = inf;
        for(int i = pre[t]; i != -1; i = pre[edge[i^1]. to])
        {
            if(Min > edge[i]. cap - edge[i]. flow)
                Min = edge[i]. cap - edge[i]. flow;
        }
        for(int i = pre[t]; i != -1; i = pre[edge[i^1]. to])
        {
            edge[i]. flow += Min;
            edge[i^1]. flow -= Min;
            cost += edge[i]. cost * Min;
        }
        flow += Min;
    }
}

int main()
{
    int u,v,w1,w2;
    int cas = 0;
    while(~scanf("%d%d",&n,&m))
    {
        init();
        for(int i = 0; i < m; i++)
        {
            scanf("%d%d%d%d",&u,&v,&w1,&w2);
            //每一条边分解成两条流量是1的边
            //一条费用是w1
            addedge(u,v,1,w1);
            //另外一条是w1+w2
            addedge(u,v,1,w1+w2);
        }
        //手动加入源点汇点 流量都是2
        addedge(0,1,2,0);
        addedge(n,n+1,2,0);
        int ans=0;
        solve(0,n+1,ans);
        printf("Case %d: %d
",++cas,ans);
    }
    return 0;
}/*

4 4
1 2 5 1
2 4 6 0
1 3 4 0
3 4 9 1
4 4
1 2 5 10
2 4 6 10
1 3 4 10
3 4 9 10

*/