BAPC 2016 ----Brexit （BFS + vector）

听说这道题和vector更配哦~

用vector存放盟友，这道题就非常简单了

题意：

一个由C个国家组成的联盟，有一天L离开了，对于任意国家，只要盟友少于之前的一半，他也会离开，问X会不会离开

思路：

用vector记录与当前国家相连的国家，num数组记录开始各个国家的盟友数目，lost记录离开联盟的盟友个数，

BFS把离开的国家跑一遍答案就出来了

# include <iostream>
# include <stdio.h>
# include <string.h>
# include <memory>
# include <cmath>
# include <string>
# include <cstdlib>
# include <queue>
# include <vector>
# include <algorithm>
# include <set>
# include <map>

using namespace std;

int num[1000005], lost[1000005], vis[1000005];
queue <int> q;
vector <int> G[1000005];

int main()
{
    int C, P, X, L;
    int x, y, v, i, flag, u;

    memset(num, 0, sizeof(num));
    memset(lost, 0, sizeof(lost));
    memset(vis, 0, sizeof(vis));

    scanf("%d %d %d %d", &C, &P, &X, &L);
    while(P--)
    {
        scanf("%d %d", &x, &y);
        G[x].push_back(y);
        G[y].push_back(x);
        num[x]++;
        num[y]++;
    }

    q.push(L);
    flag = 1;
    vis[L] = 1;  //开始没标记L，wa了一下午
    while(!q.empty())
    {
        u = q.front();
        q.pop();
        if(u==X)
        {
            flag = 0;
            printf("leave
");
            break;
        }
        else
        {
            int NN = G[u].size();
            for(i=0; i<NN; i++)
            {
                v = G[u][i];
                lost[v]++;
                if(lost[v]*2>=num[v]&&vis[v]==0)
                {
                    q.push(v);
                    vis[v] = 1;
                }
            }
        }

    }
    if(flag==1) printf("stay
");
    return 0;
}