CCF_ 201403-4_无线网络

分散点的bfs，先建立一个互相是否可达的二维数组，vis[i][j]代表到第i个点，走了j步的状态，注意判断新增路由器数量是否超过K。

#include<cstdio>
#include<iostream>
#include<queue>
using namespace std;

struct point{
    long long x,y,num,counts,sets;
}a[205];

int cango[205][205] = {0},vis[205][205] = {0};

int main()
{
    int n,m,k;
    long long r;
    cin >> n >> m >> k >> r;
    for(int i = 1;i <= m+n;i++)
    {
       cin >> a[i].x >> a[i].y;
       a[i].num = i;
    }
    for(int i = 1;i < m+n;i++)
    {
        for(int j = i;j <= m+n;j++)
        {
            if((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y) <= r*r)
            {
                cango[i][j] = cango[j][i] = 1;
            }
        }
    }
    a[1].counts = 0;
    a[1].sets = 0;
    vis[1][0] = 1;
    queue<point> q;
    q.push(a[1]);
    while(!q.empty())
    {
        long long num = q.front().num,counts = q.front().counts;
        q.pop();
        if(num == 2)
        {
            cout << counts-1 << endl;
            return 0;
        }
        for(int i = 1;i <= m+n;i++)
        {
            long long sets;
            if(i > n)    sets = q.front().sets+1;
            else        sets = q.front().sets;
            if(cango[num][i] && !vis[i][counts+1] && sets <= k)
            {
                a[i].sets = sets;
                a[i].counts = counts+1;
                q.push(a[i]);
                vis[i][counts+1] = 1;
            }
        }
    }
}