poj 2186 Popular Cows(Kosaraju + 变形)

感想详见：http://blog.csdn.net/zxy_snow/archive/2011/01/18/6149058.aspx

 

我用了两种做法。

 

Kosaraju 缩点后加DFS，搜索能满足题意的点，也就是缩点后集合能到达其他任意集合的点。1800+MS。

 

Kosaraju 缩点后查找入度为0的点，如果只有1个，那么这个肯定就是答案。250MS。

 

可以证明一下，如果没有入度为0的点，那么缩点肯定是一个环，这样的话，求强连通分量的时候，肯定是当成一个求的，就不会分这么多个。

 

如果入度为0的点不为1，那么肯定有不能到达的缩点，那么肯定这个集合的缩点都不能被其他的缩点到达，所以肯定这个集合的cows不喜欢其他的cows。

 

呃。我的缩点是在转置后的邻接表基础上建成的。

 

如果是转置前的，需要找出度为0的点。

 

DFS

 

#include <stdio.h>#include <stdlib.h>#include <iostream>#include <string.h>#include <stack>#define MAX 10010using namespace std;typedef struct cow{int to;cow *next;}cow;typedef struct shrdot{int num,head;shrdot *next;}shrdot;shrdot *shr[MAX],mall[MAX*5];cow *c[MAX],*ce[MAX],node[MAX*10];stack <int> fin;int n,m,cou,sou,ind,sum;int visit[MAX],pre[MAX],tmp[MAX],used[MAX];void init(){memset(c,'/0',sizeof(c));memset(ce,'/0',sizeof(ce));memset(node,'/0',sizeof(node));memset(mall,'/0',sizeof(mall));memset(shr,'/0',sizeof(shr));cou = 0;}void input(){int from,to;scanf("%d%d",&n,&m);while(m--){scanf("%d%d",&from,&to);node[cou].to = to;node[cou].next = c[from];c[from] = &node[cou++];}}void Transpose(cow **c,cow **ce){int i;cow *head;for(i=1; i<=n; i++){head = c[i];while( head != NULL ){node[cou].to = i;node[cou].next = ce[head->to];ce[head->to] = &node[cou++];head = head->next;}}}void FirDFS(cow **c,int s){visit[s] = 1;cow *head = c[s];while( head != NULL ){if( !visit[head->to] )FirDFS(c,head->to);head = head->next;}fin.push(s);}void SecDFS(cow **ce,int s,int ss){visit[s] = 1;cow *head = ce[s];while( head != NULL ){if( !visit[head->to] ){sou++;pre[head->to] = ss;SecDFS(ce,head->to,ss);}elsetmp[ind++] = head->to;head = head->next;}}void ShrDFS(int s){visit[s] = 1;shrdot *head = shr[s];while( head != NULL ){if( !visit[head->head] ){sum += shr[head->head]->num;ShrDFS(head->head);}head = head->next;}}void Kosaraju(cow **c, cow **ce){int i,num,k,ans;Transpose(c,ce);memset(visit,0,sizeof(visit));for(i=1; i<=n; i++)if( c[i] != NULL && !visit[i] )FirDFS(c,i);memset(visit,0,sizeof(visit));memset(pre,0,sizeof(pre));int shrcou = 0,jilu = 0;while( !fin.empty() ){num = fin.top();fin.pop();ind = 0; sou = 1; memset(used,0,sizeof(used));if( !visit[num] ){pre[num] = shrcou;SecDFS(ce,num,shrcou);mall[jilu].num = sou;mall[jilu].head = shrcou;mall[jilu].next = shr[shrcou];shr[shrcou] = &mall[jilu++]; for(i=0; i<ind; i++)if( tmp[i] != num && pre[tmp[i]] != shrcou && !used[pre[tmp[i]]] ){mall[jilu].num = shr[pre[tmp[i]]]->num;mall[jilu].head = shr[pre[tmp[i]]]->head;mall[jilu].next = shr[shrcou]->next;shr[shrcou]->next = &mall[jilu++];used[pre[tmp[i]]] = 1;}shrcou++;}}ans = 0;for(i=0; i<shrcou; i++)// 第三个DFS，搜满足题意的缩点。 {sum = shr[i]->num;memset(visit,0,sizeof(visit));ShrDFS(i);if( sum == n ) ans += shr[i]->num;}printf("%d/n",ans);}int main(){init();input();Kosaraju(c,ce);return 0;} 

 

找入度

 

#include <stdio.h>#include <stdlib.h>#include <iostream>#include <string.h>#include <stack>#define MAX 10010using namespace std;typedef struct cow{int to;cow *next;}cow;cow *c[MAX],*ce[MAX],node[MAX*10];stack <int> fin;int n,m,cou,sou,ind,sum;int visit[MAX],pre[MAX],tmp[MAX],used[MAX];void init(){memset(c,'/0',sizeof(c));memset(ce,'/0',sizeof(ce));memset(node,'/0',sizeof(node));cou = 0;}void input(){int from,to;scanf("%d%d",&n,&m);while(m--){scanf("%d%d",&from,&to);node[cou].to = to;node[cou].next = c[from];c[from] = &node[cou++];}}void Transpose(cow **c,cow **ce){int i;cow *head;for(i=1; i<=n; i++){head = c[i];while( head != NULL ){node[cou].to = i;node[cou].next = ce[head->to];ce[head->to] = &node[cou++];head = head->next;}}}void FirDFS(cow **c,int s){visit[s] = 1;cow *head = c[s];while( head != NULL ){if( !visit[head->to] )FirDFS(c,head->to);head = head->next;}fin.push(s);}void SecDFS(cow **ce,int s,int ss){visit[s] = 1;cow *head = ce[s];while( head != NULL ){if( !visit[head->to] ){sou++;pre[head->to] = ss;SecDFS(ce,head->to,ss);}elsetmp[ind++] = head->to;head = head->next;}}void Kosaraju(cow **c, cow **ce){int i,num,count,tmpi;int in[MAX],setn[MAX];Transpose(c,ce);memset(visit,0,sizeof(visit));for(i=1; i<=n; i++)if( c[i] != NULL && !visit[i] )FirDFS(c,i);memset(visit,0,sizeof(visit));memset(pre,0,sizeof(pre));memset(in,0,sizeof(in));memset(setn,0,sizeof(setn));while( !fin.empty() ){num = fin.top();fin.pop();ind = 0; sou = 1; memset(used,0,sizeof(used));if( !visit[num] ){pre[num] = num;SecDFS(ce,num,num);setn[num] = sou;used[num] = 1;for(i=0; i<ind; i++)if( !used[pre[tmp[i]]] ){in[pre[tmp[i]]]++; //记录点的入度。 used[pre[tmp[i]]] = 1;}}}for(i=1; i<=n; i++)//如果有点一直没有被访问到，肯定是单独的点，输出0 if( !visit[i] ){printf("0/n");return ;}count = 0;for(i=1; i<=n; i++)if( pre[i] == i && in[i] == 0 ) //找入度为0的点 {tmpi = i;count++;}if( count == 1 ) // 如果入度为0的点只有一个，那么肯定就是满足题意的，其他都不对。 printf("%d/n",setn[tmpi]);elseprintf("0/n");}int main(){init();input();Kosaraju(c,ce);return 0;}