Gym - 101875I I Will Go (dfs序)

题意:N个人要参加一个局,每个人有自己的好朋友,如果他的好朋友来,他才有可能来。N个人的关系不够成环。Q次查询,问若x来了,y是否肯定来。
分析:若点y是x的祖先,则y肯定回来。一次dfs确定每个点覆盖的区间,若点x的dfs序在y的覆盖区间内,则y肯定会来。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5+5;
typedef long long LL;
const int INF = 0x3f3f3f3f;
struct Edge{
    int v,next;
}edges[maxn];
int tot,head[maxn],dfn;
int L[maxn],R[maxn];
int ind[maxn];
void init()
{
    memset(head,-1,sizeof(head));
    memset(ind,0,sizeof(ind));
    memset(L,0,sizeof(L));
    tot= dfn = 0;
}

void AddEdge(int u,int v)
{
    edges[tot] = (Edge){v,head[u]};
    head[u] = tot++;
}

void dfs(int u)
{
    L[u] = ++dfn;
    for(int i=head[u];~i;i=edges[i].next){
        int v = edges[i].v;
        dfs(v);
    }
    R[u] = dfn;
}

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    int st,N,M;
    int u,v,tmp;
    while(scanf("%d %d",&N,&M)==2){
        init();
        for(int i =1;i<=N;++i){
            scanf("%d",&u); 
            if(u==-1) continue;
            u++;
            AddEdge(u,i);
            ind[i]++;
        }
        for(int i=1;i<=N;++i){
            if(!ind[i]) dfs(i);
        }
        while(M--){
            scanf("%d %d",&u,&v);
            u++,v++;
            //cout<<L[v]<<" "<<R[v]<<endl;
            if(L[u]>=L[v] && L[u]<=R[v]) printf("Yes\n");
            else printf("No\n");
        }
    }
    return 0;
}

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转载自www.cnblogs.com/xiuwenli/p/9574711.html
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