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;
}