Popular Cows
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 28379 | Accepted: 11488 |
Description
Every cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M <= 50,000) ordered pairs of the form (A, B) that tell you that cow A thinks that cow B is popular. Since popularity is transitive, if A thinks B is popular and B thinks C is popular, then A will also think that C is
popular, even if this is not explicitly specified by an ordered pair in the input. Your task is to compute the number of cows that are considered popular by every other cow.
popular, even if this is not explicitly specified by an ordered pair in the input. Your task is to compute the number of cows that are considered popular by every other cow.
Input
* Line 1: Two space-separated integers, N and M
* Lines 2..1+M: Two space-separated numbers A and B, meaning that A thinks B is popular.
* Lines 2..1+M: Two space-separated numbers A and B, meaning that A thinks B is popular.
Output
* Line 1: A single integer that is the number of cows who are considered popular by every other cow.
Sample Input
3 3 1 2 2 1 2 3
Sample Output
1
题意:A认为B受欢迎,B认为C受欢迎,那么A就认为C受欢迎。给定N头牛,M个认为受欢迎情况。确定受所有牛欢迎的这些牛的数目。
思路:首先判断有向图是否连通,若不连通则答案为0。若连通则进行缩点（注意有向图与无向图的缩点的不同之处）,缩点之后得到一个有向树。若树中存在多个叶子结点，则答案为0，否则答案为 缩成叶子结点的那个连通分量重结点的数目。
下面是第二种有向图缩点方法,一次dfs,边反向后再进行一次rdfs.解释见代码
#include"cstdio" #include"cstring" #include"vector" using namespace std; const int MAXN=10005; int V,E; vector<int> G[MAXN]; vector<int> rG[MAXN]; vector<int> vs; bool used[MAXN]; int cmp[MAXN]; void add_edge(int u,int v) { G[u].push_back(v); rG[v].push_back(u); } void dfs(int u) { used[u]=true; for(int i=0;i<G[u].size();i++) if(!used[G[u][i]]) dfs(G[u][i]);//假设u所能访问的结点都处于同一连通分量 vs.push_back(u); //后续遍历 } void rdfs(int u,int k) { used[u]=true; cmp[u]=k; for(int i=0;i<rG[u].size();i++) if(!used[rG[u][i]]) rdfs(rG[u][i],k); } int scc() { memset(used,false,sizeof(used)); for(int i=1;i<=V;i++) if(!used[i]) dfs(i); memset(used,false,sizeof(used)); int k=0; for(int i=vs.size()-1;i>=0;i--)//倒着访问,保证边反向后各个连通分量互不影响 if(!used[vs[i]]) rdfs(vs[i],k++);//对应后续遍历,若边反向后,起点u仍能遍历整个连通分量,那么它们处以同一连通分量中.否则处于不同的连通分量 return k; } int deg[MAXN]; int seek() { memset(deg,0,sizeof(deg)); int k=scc(); for(int i=1;i<=V;i++) for(int j=0;j<G[i].size();j++) { int to=G[i][j]; if(cmp[i]!=cmp[to]) { deg[cmp[i]]++; } } int v=-1; int flag=0; int ans=0; for(int i=1;i<=V;i++) if(deg[cmp[i]]==0) { ans++; if(cmp[i]!=v) { v=cmp[i]; flag++; } if(flag>1) return 0; } return ans; } int main() { while(scanf("%d%d",&V,&E)!=EOF) { for(int i=1;i<=V;i++) { vs.clear(); G[i].clear(); rG[i].clear(); cmp[i]=0; } for(int i=0;i<E;i++) { int u,v; scanf("%d%d",&u,&v); add_edge(u,v); } int ans=seek(); printf("%d ",ans); } return 0; }
标准tarjan算法对有向图缩点。
#include<cstdio> #include<cstring> #include<algorithm> #include<vector> using namespace std; const int MAXN=10005; vector<int> mp[MAXN]; int n,m; int dfn[MAXN],low[MAXN],time; int stack[MAXN],top; bool ins[MAXN]; int belong[MAXN],cnt; void dfs(int u) { dfn[u]=low[u]=++time; stack[top++]=u; ins[u]=true; for(int i=0;i<mp[u].size();i++) { int v=mp[u][i]; if(!dfn[v]) { dfs(v); low[u]=min(low[u],low[v]); } else if(ins[v]) low[u]=min(low[u],dfn[v]); } if(dfn[u]==low[u]) { cnt++; int v; do{ v=stack[--top]; belong[v]=cnt; ins[v]=false; }while(u!=v); } } int deg[MAXN]; void cal() { memset(deg,0,sizeof(deg)); int res=0; for(int i=1;i<=n;i++) for(int j=0;j<mp[i].size();j++) { int v=mp[i][j]; if(belong[i]!=belong[v]) { deg[belong[i]]++; } } int mark; for(int i=1;i<=cnt;i++) if(deg[i]==0) { mark=i; res++; } if(res!=1) { printf("0 "); return; } res=0; for(int i=1;i<=n;i++) if(belong[i]==mark) res++; printf("%d ",res); } int main() { while(scanf("%d%d",&n,&m)!=EOF) { for(int i=1;i<=n;i++) mp[i].clear(); memset(dfn,0,sizeof(dfn)); memset(low,0,sizeof(low)); time=0; top=0; cnt=0; memset(ins,false,sizeof(ins)); for(int i=0;i<m;i++) { int u,v; scanf("%d%d",&u,&v); mp[u].push_back(v); } for(int i=1;i<=n;i++) if(!dfn[i]) dfs(i); cal(); } return 0; }