bzoj 1051: [HAOI2006]受欢迎的牛

tarjan强连通分量求缩点重构图，出度为0的点若只有一个则输出其代表强连通分量的大小，否则无解。

因为一旦出度为0就没人被他认为长得帅

  1 /*
  2 ID:WULALA
  3 PROB:bzoj1051 
  4 LANG:C++
  5 */
  6 #include <cstdio>
  7 #include <cstring>
  8 #include <algorithm>
  9 #include <cmath>
 10 #include <iostream>
 11 #include <fstream>
 12 #include <ctime>
 13 #define N 10008
 14 #define M 50008
 15 #define mod
 16 #define mid(l,r) ((l+r) >> 1)
 17 #define INF 0x7ffffff
 18 using namespace std;
 19 
 20 int n,m,dfn[N],low[N],que[N],head[N],cnt,h[N],scc,belong[N],ans,hav[N],r;
 21 bool vis[N],inq[N],flag;
 22 
 23 struct WULALA
 24 {
 25     int node,next;
 26 }e[M],d[M];
 27 
 28 void init()
 29 {
 30     scanf("%d%d",&n,&m);
 31     for (int i = 1;i <= m;i++)
 32     {
 33         int x,y;
 34         scanf("%d%d",&x,&y);
 35         e[i].node = y;
 36         e[i].next = head[x];
 37         head[x] = i;
 38     }
 39 }
 40 
 41 void dfs(int a)
 42 {
 43     vis[a] = true;
 44     low[a] = dfn[a] = ++cnt;
 45     inq[a] = true; que[++r] = a;
 46     int c = head[a];
 47     while(c)
 48     {
 49         if (!vis[e[c].node]) 
 50         {
 51             dfs(e[c].node);
 52             low[a] = min(low[a],low[e[c].node]);
 53         }
 54         else if (inq[e[c].node]) low[a] = min(low[a],dfn[e[c].node]);//要判是否在队列里！！ 
 55         c = e[c].next;
 56     }
 57     if (low[a] == dfn[a])
 58     {
 59         belong[a] = ++scc;
 60         hav[scc] = 1;
 61         while (que[r] != a)
 62         {
 63             belong[que[r]] = scc;
 64             inq[que[r]] = false;
 65             ++hav[scc];
 66             --r;
 67         }
 68         inq[que[r]] = false;//!!!
 69         --r;//!!!
 70     }
 71 }
 72 
 73 void rebuild()
 74 {
 75     cnt = 0;
 76     for (int i = 1;i <= n;i++)
 77     {
 78         int c = head[i];
 79         while(c)
 80         {
 81             if (belong[i] != belong[e[c].node])
 82             {
 83                 d[++cnt].node = belong[e[c].node];    
 84                 d[cnt].next = h[belong[i]];
 85                 h[belong[i]] = cnt;
 86             }
 87             c = e[c].next;
 88         }
 89     }
 90 }
 91 
 92 void tarjan()
 93 {
 94 
 95     for (int i = 1;i <= n;i++)
 96         if (!vis[i]) dfs(i);
 97     rebuild();
 98 }
 99 
100 void work()
101 {
102     for (int i = 1;i <= scc;i++)
103         if (!h[i])
104         {
105             if (ans)
106             {
107                 ans = 0;
108                 return;
109             }
110             else ans = hav[i];
111         }
112 }
113 
114 int main()
115 {
116     init();
117     tarjan();
118     work();
119     printf("%d\n",ans);
120     return 0;
121 }
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posted @ 2014-01-06 18:50  乌拉拉979  阅读(182)  评论(0)  编辑  收藏  举报