UVALive - 6266 Admiral 费用流

UVALive - 6266 Admiral

题意：找两条完全不相交不重复的路使得权值和最小。

思路：比赛的时候时间都卡在D题了，没有仔细的想这题，其实还是很简单的，将每个点拆开，连一条容量为1，费用为0的边，起点和终点容量为2，两点之间有边就加一条容量为1，费用为权值的边，这样跑一边费用流就可以了。

 #pragma comment(linker, "/STACK:1000000000")
 #include <bits/stdc++.h>
 #define LL long long
 #define INF 0x3f3f3f3f
 #define IN freopen("in.txt","r",stdin);
 #define OUT freopen("out.txt","w",stdout);
 using namespace std;
 #define MAXN 9999
 #define V 2005
 #define E 30005
 int vis[V];
 int dist[V];
 int pre[V];

 struct Edge{
     int u,v,c,cost,next;
 }edge[E];
 int head[V],cnt;

 void init(){
     cnt=;
     memset(head,-,sizeof(head));
 }
 void addedge(int u,int v,int c,int cost)
 {
     edge[cnt].u=u;edge[cnt].v=v;edge[cnt].cost=cost;
     edge[cnt].c=c;edge[cnt].next=head[u];head[u]=cnt++;

     edge[cnt].u=v;edge[cnt].v=u;edge[cnt].cost=-cost;
     edge[cnt].c=;edge[cnt].next=head[v];head[v]=cnt++;
 }

 bool spfa(int begin,int end){
     int u,v;
     queue<int> q;
     for(int i=;i<=end+;i++){
         pre[i] = -;
         vis[i] = ;
         dist[i] = INF;
     }
     vis[begin]=;
     dist[begin]=;
     q.push(begin);
     while(!q.empty()){
         u=q.front();
         q.pop();
         vis[u]=;
         for(int i=head[u];i!=-;i=edge[i].next){
             if(edge[i].c>){
                 v=edge[i].v;
                 if(dist[v]>dist[u]+edge[i].cost){
                     dist[v]=dist[u]+edge[i].cost;
                     pre[v]=i;
                     if(!vis[v]){
                         vis[v]=true;
                         q.push(v);
                     }
                 }
             }
         }
     }
     return dist[end] != INF;
 }

 int MCMF(int begin,int end){
     int ans=,flow;
     int flow_sum=;
     while(spfa(begin,end)){
         flow=INF;
         for(int i=pre[end];i!=-;i=pre[edge[i].u])
             if(edge[i].c<flow)
                 flow=edge[i].c;
         for(int i=pre[end];i!=-;i=pre[edge[i].u]){
             edge[i].c-=flow;
             edge[i^].c+=flow;
         }
         ans+=dist[end];
         flow_sum += flow;
     }
     //cout << flow_sum << endl;
     return ans;
 }

 int main()
 {
     //IN;
     int n, m, x, y, z;
     while(~scanf("%d%d", &n, &m)){
         init();
         int s = , t =  * n;
         for(int i = ; i < n; i++){
             addedge(i, i + n, , );
         }
         addedge(, n + , , );
         addedge(n, n << , , );
         for(int i = ; i <= m; i++){
             scanf("%d%d%d", &x, &y, &z);
             addedge(x + n, y, , z);
         }
         int ans = MCMF(s, t);
         printf("%d\n", ans);
     }
     return ;
 }