ZJNU 2133

将辈分差距转为数字

例如 A 是 B son

A=B-1

A 是 B grandfather

A=B+2
然后编号1数字设置为0，建图bfs

最后搜索编号2到100是否存在>0的数即可

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int> P;
vector<P> v[105];
queue<int> q;
int dis[105];
bool vis[105];
int main(){
    memset(dis,0x3f,sizeof dis);
    int N,i,now,cnt,a,b;
    char s[15];
    scanf("%d",&N);
    while(N--){
        scanf("%d is %d's %s",&a,&b,s);
        if(strcmp(s,"father.")==0){
            v[a].push_back(P(b,-1));
            v[b].push_back(P(a,1));
        }
        else if(strcmp(s,"son.")==0){
            v[a].push_back(P(b,1));
            v[b].push_back(P(a,-1));
        }
        else if(strcmp(s,"brother.")==0){
            v[a].push_back(P(b,0));
            v[b].push_back(P(a,0));
        }
        else if(strcmp(s,"grandson.")==0){
            v[a].push_back(P(b,2));
            v[b].push_back(P(a,-2));
        }
        else if(strcmp(s,"grandfather.")==0){
            v[a].push_back(P(b,-2));
            v[b].push_back(P(a,2));
        }
    }
    vis[1]=true;
    dis[1]=0;
    q.push(1);
    while(!q.empty()){
        now=q.front();
        q.pop();
        cnt=v[now].size();
        for(i=0;i<cnt;i++)
            if(!vis[v[now][i].first]){
                vis[v[now][i].first]=true;
                dis[v[now][i].first]=dis[now]+v[now][i].second;
                q.push(v[now][i].first);
            }
    }
    bool flag=true;
    for(i=2;i<=100;i++)
        if(dis[i]!=0x3f3f3f3f&&dis[i]>0){
            flag=false;
            break;
        }
    puts(flag?"2333333...":"ku jiu ru hou xin zuo tong.");
    
    return 0;
}