CSU 1506: Double Shortest Paths（最小费用最大流）

题目链接：http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1506

Description

Input

There will be at most 200 test cases. Each case begins with two integers n, m (1<=n<=500, 1<=m<=2000), the number of caves and passages. Each of the following m lines contains four integers u, v, di and ai (1<=u,v<=n, 1<=di<=1000, 0<=ai<=1000). Note that there can be multiple passages connecting the same pair of caves, and even passages connecting a cave and itself.

Output

For each test case, print the case number and the minimal total difficulty.

Sample Input

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

Sample Output

Case 1: 23
Case 2: 24

HINT

Source

湖南省第十届大学生计算机程序设计竞赛

题意：

有两个人；

给出路径之间第一个人走所需要的费用和第二个人走所需要的费用（在第一个人所需的 费用上再加上第二次的费用）；

求两个人一共所需要的最小费用！

代码如下：（套用kuangbin模板）

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
//点的总数为 N，点的编号 1~N
const int MAXN = 10000;
const int MAXM = 100000;
const int INF = 0x3f3f3f3f;
struct Edge
{
    int to, next, cap, flow, cost;
} edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N, M;
void init(int n)
{
    N = n;
    tol = 0;
    memset(head, -1, sizeof(head));
}
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;
    for(int i = 0; i < N; i++)
    {
        dis[i] = INF;
        vis[i] = false;
        pre[i] = -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;
}
//返回的是最大流， cost存的是最小费用
int minCostMaxflow(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;
    }
    return flow;
}
int main()
{
    int n, m;
    int u, v, cost1, cost2;
    int cas = 0;
    while(~scanf("%d%d",&n,&m))
    {
        init(n+2);//注意
        for(int i = 0; i < m; i++)
        {
            scanf("%d%d%d%d",&u,&v,&cost1,&cost2);
            addedge(u,v,1,cost1);//流量为1
            addedge(u,v,1,cost1+cost2);
        }
        //添加一个总的起点，一个终点，使编号从0~n
        addedge(0,1,2,0);//流量为2,费用为0
        addedge(n,n+1,2,0);
        int ans = 0;
        minCostMaxflow(0,n+1,ans);
        printf("Case %d: %d\n",++cas,ans);
    }
    return 0;
}