HDU 4435

思路就是贪心，i从n枚举到2，依次判断如果[0,i-1]全设为加油站是否可行，这里用一个bfs即可实现，总复杂度o（n^3）

 

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

using namespace std;

int n,d;

struct Point{

    int x,y;

}point[200];

int dis[200][200];

bool vis[200],ok[200];

int dd[200];



void bfs(){

    queue<int>q;

    q.push(1);

    memset(ok,0,sizeof(ok));

    ok[1]=1;

    for(int i=1;i<=n;i++)dd[i]=1000000;

    while(!q.empty()){

        int u=q.front();

        q.pop();

        for(int i=1;i<=n;i++){

            if(i==u)continue;

            if(dis[u][i]<=d && ok[i]==0){

                if(vis[i]) q.push(i),ok[i]=1;

                else dd[i]=min(dd[i],dis[u][i]);

            }

        }

    }

}



void solve(){

    int i,j;

    for(i=1;i<=n;i++)

        vis[i]=1;

    for(i=n;i>1;i--){

        vis[i]=0;

        bfs();

        bool flag=1;

        for(j=1;j<=n;j++){

            if(vis[j]){

                if(ok[j]==0){

                    flag=0;break;

                }

            }

            else if(dd[j]*2>d){

                flag=0;break;

            }

        }

        if(!flag) vis[i]=1;

    }

}



int main(){

    int i,j;

    int tem;

    while(scanf("%d %d",&n,&d)!=EOF){

        for(i=1;i<=n;i++) scanf("%d %d",&point[i].x,&point[i].y);

        bool flag=1;

        for(i=1;i<=n;i++)

            for(j=i+1;j<=n;j++)

                dis[i][j]=dis[j][i]=ceil(sqrt((point[i].x-point[j].x)*(point[i].x-point[j].x)+(point[i].y-point[j].y)*(point[i].y-point[j].y)));

        for(i=1;i<=n;i++){

            tem=1000000;

            for(j=1;j<=n;j++){

                if(i==j)continue;

                if(dis[i][j]<tem) tem=dis[i][j];

            }

            if(tem>d) flag=0;

        }

        if(flag==0){

            printf("-1\n");

            continue;

        }

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

        solve();

        j=n;

        while(1){

            if(vis[j]==0)

                j--;

            else break;

        }

        for(;j>=1;j--) printf("%d",vis[j]);

        puts("");

    }

    return 0;

}