poj3114Countries in War（缩点+DIJK）

http://poj.org/problem?id=3114

缩点+DIJK 注意缩点之后有重边啊 floyd会TLE

  1 #include <iostream>
  2 #include<cstring>
  3 #include<cstdio>
  4 #include<stdlib.h>
  5 #include<algorithm>
  6 #include<stack>
  7 #define N 510
  8 #define M 301000
  9 #define INF 0xfffffff
 10 using namespace std;
 11 struct node
 12 {
 13     int u,v,next,w;
 14 }edge[M];
 15 int t,low[N],pre[N],sccno[N],head[N],scc,dep,ww[N][N],vis[N],dis[N];
 16 int o[N][N];
 17 void init()
 18 {
 19     t = 0;
 20     memset(head,-1,sizeof(head));
 21 }
 22 void add(int u,int v,int w)
 23 {
 24     edge[t].u  =u;
 25     edge[t].v = v;
 26     edge[t].w = w;
 27     edge[t].next = head[u];
 28     head[u] = t++;
 29 }
 30 stack<int>s;
 31 void dfs(int u)
 32 {
 33     low[u] = pre[u] = ++dep;
 34     s.push(u);
 35     for(int i = head[u] ; i!=-1 ; i = edge[i].next)
 36     {
 37         int v = edge[i].v;
 38         if(!pre[v])
 39         {
 40             dfs(v);
 41             low[u] = min(low[u],low[v]);
 42         }
 43         else if(!sccno[v])
 44         low[u] = min(low[u],pre[v]);
 45     }
 46     if(low[u]==pre[u])
 47     {
 48         scc++;
 49         for(;;)
 50         {
 51             int x = s.top();s.pop();
 52             sccno[x] = scc;
 53             if(x==u)break;
 54         }
 55     }
 56 }
 57 void find_scc(int n)
 58 {
 59     scc=0;dep=0;
 60     memset(low,0,sizeof(low));
 61     memset(pre,0,sizeof(pre));
 62     memset(sccno,0,sizeof(sccno));
 63     for(int i = 1; i <= n ; i++)
 64     if(!pre[i])
 65     dfs(i);
 66 }
 67 int dijk(int st,int en,int n)
 68 {
 69     int i,j,k,mi;
 70     memset(vis,0,sizeof(vis));
 71     for(i = 1; i <= n ; i++)
 72     dis[i] = o[st][i];
 73     dis[st] = 0;
 74     for(i = 1 ;i <= n ; i++)
 75     {
 76         mi = INF;
 77         for(j = 1; j <= n ; j++)
 78             if(!vis[j]&&dis[j]<=mi)
 79             mi = dis[k=j];
 80         vis[k] = 1;
 81         for(j = 1; j <= n ; j++)
 82         if(dis[j]>o[k][j]+dis[k])
 83         dis[j] = o[k][j]+dis[k];
 84     }
 85     return dis[en];
 86 }
 87 int main()
 88 {
 89     int i,j,n,m,k,a,b,c,g;
 90     while(cin>>n>>m)
 91     {
 92         if(n==0&&m==0)
 93         break;
 94         memset(ww,0,sizeof(ww));
 95         init();
 96         for(i =1; i <= m ;i++)
 97         {
 98             scanf("%d%d%d",&a,&b,&c);
 99             add(a,b,c);
100         }
101         find_scc(n);
102         for(i = 1; i <= n ; i++)
103            for(j = 1; j <= n ;j++)
104              o[i][j] = INF;
105         for(i = 1; i <= scc ; i++)
106         o[i][i] = 0;
107         for(i = 0; i < t ; i++)
108         {
109             int u = edge[i].u,v = edge[i].v;
110             if(sccno[u]!=sccno[v])
111             {
112                 o[sccno[u]][sccno[v]] = min(o[sccno[u]][sccno[v]],edge[i].w);//没算重的之前 WA惨了 
113             }
114         }
115         scanf("%d",&k);
116         while(k--)
117         {
118             scanf("%d%d",&a,&b);
119             int xx = dijk(sccno[a],sccno[b],scc);
120             if(xx==INF)
121             printf("Nao e possivel entregar a carta\n");
122             else
123             cout<<xx<<endl;
124         }
125         puts("");
126     }
127     return 0;
128 }
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posted @ 2013-06-19 12:36  _雨  阅读(273)  评论(0)  编辑  收藏  举报