UVA 1658 Admiral 最小费用最大流

题目链接：

　　　　http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=51253

题意：

给出一个v个点e条边的有向加权图,求1~v的两条不相交的路径使得权值和最小

题解：

最小费用最大流

吧2到v-1的每隔结点拆点x和x'，中间连一条容量为1费用为0的边，然后求得是1到v流量为2的最小费用流

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn = 1e6+20, M = 30005, mod = 1e9+7, inf = 0x3f3f3f3f;
typedef long long ll;
//不同为1，相同为0

const int MAXN = 100000;
const int MAXM = 1000000;
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;//节点总个数，节点编号从0~N-1
void init(int n)
{
    N = n;
    tol = 0;
    memset(head,-1,sizeof(head));
}
void add(int u,int v,int cap,int cost)  //点u至点v,容量，花费
{
    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(flow<2&&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;
        while(scanf("%d%d",&n,&m)!=EOF) {
        int S = 0, T = n*2;
        init (5000);
        for(int i=1;i<=m;i++) {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
          if(a!=1&&a!=n)
            add(a+n,b,1,c);
            else add(a,b,1,c);
        }
        add(S,1,2,0);add(n,T,2,0);
        for(int i=2;i<n;i++) add(i,i+n,1,0);
        int ans = 0;
        minCostMaxflow(S,T,ans);
        printf("%d
",ans);
        }

    return 0;
}