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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