nyoj 120 校园网络
题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=120
思路:先将原图强连通缩点为新图,统计新图中入度,出度为0的点的个数,两者取最大值即为答案。
代码如下:
#include "stdio.h" #include "string.h" #include "stack" #include "vector" using namespace std; #define N 105 #define M 10005 struct node { int x,y; int next; }edge[2*M]; int idx,head[N]; void Init(){ idx=0; memset(head,-1,sizeof(head)); } inline int MIN(int a,int b){ return a<b?a:b; } inline int MAX(int a,int b){ return a>b?a:b; } void Add(int x,int y) { edge[idx].x=x; edge[idx].y=y; edge[idx].next=head[x]; head[x]=idx++; } bool vis[N]; int dfn[N],low[N]; int dfs_clock; int countt; int belong[N]; stack<int> q; void DFS(int x) //强连通缩点算法 { int i,y; q.push(x); vis[x] = true; dfn[x] = low[x] = ++dfs_clock; for(i=head[x]; i!=-1; i=edge[i].next) { y = edge[i].y; if(!dfn[y]) { DFS(y); low[x] = MIN(low[x],low[y]); } else if(vis[y]) low[x] = MIN(dfn[y],low[x]); } if(low[x]==dfn[x]) { int temp; countt++; while(1) { temp = q.top(); q.pop(); belong[temp] = countt; vis[temp] = false; if(temp==x) break; } } } int ru_du[N],chu_du[N]; int Solve(int n) { int i; int x,y; dfs_clock = countt = 0; memset(dfn,0,sizeof(dfn)); memset(vis,false,sizeof(vis)); memset(belong,0,sizeof(belong)); for(i=1; i<=n; ++i) { if(!dfn[i]) DFS(i); } if(countt==2) return 0; //该图原本就是强连通图,return 0; memset(ru_du,0,sizeof(ru_du)); memset(chu_du,0,sizeof(chu_du)); for(i=0; i<idx; ++i) { x = belong[edge[i].x]; y = belong[edge[i].y]; if(x!=y) //统计入度,出度 { chu_du[x]++; ru_du[y]++; } } int ru_0=0,chu_0=0; for(i=1; i<countt; ++i) { if(ru_du[i]==0) ru_0++; if(chu_du[i]==0) chu_0++; } int ans = MAX(ru_0,chu_0); //return 入度,出度最大的那个数 return ans; } int main() { int T; int n; int i,k; scanf("%d",&T); while(T--) { Init(); scanf("%d",&n); for(i=1; i<=n; ++i) { while(scanf("%d",&k),k) Add(i,k); } printf("%d\n",Solve(n)); } return 0; }