首页 > 代码库 > hdu 3832 Earth Hour （最短路变形）

hdu 3832 Earth Hour （最短路变形）

2024-07-16 22:43:43   222人阅读

Earth Hour

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 1516    Accepted Submission(s): 606


Problem Description
Earth Hour is an annual international event created by the WWF (World Wide Fund for Nature/World Wildlife Fund), held on the last Saturday of March, that asks households and businesses to turn off their non-essential lights and electrical appliances for one hour to raise awareness towards the need to take action on climate change. 
To respond to the event of this year, the manager of Hunan University campus decides to turn off some street lights at night. Each street light can be viewed as a point in a plane, which casts flash in a circular area with certain radius.
What‘s more, if two illuminated circles share one intersection or a point, they can be regarded as connected.
Now the manager wants to turn off as many lights as possible, guaranteeing that the illuminated area of the library, the study room and the dormitory are still connected(directly or indirectly). So, at least the lights in these three places will not be turned off.
 

Input
The first line contains a single integer T, which tells you there are T cases followed.
In each case:
The first line is an integer N( 3<=N<=200 ), means there are N street lights at total.
Then there are N lines: each line contain 3 integers, X,Y,R,( 1<=X,Y,R<=1000 ), means the light in position(X,Y) can illuminate a circle area with the radius of R. Note that the 1st of the N lines is corresponding to the library, the 2nd line is corresponding to the study room, and the 3rd line is corresponding to the dorm.
 

Output
One case per line, output the maximal number of lights that can be turned off.
Note that if none of the lights is turned off and the three places are still not connected. Just output -1.
 

Sample Input
3
5
1 1 1
1 4 1
4 1 1
2 2 1
3 3 1
7
1 1 1
4 1 1
2 4 1
1 3 1
3 1 1
3 3 1
4 3 1
6
1 1 1
5 1 1
5 5 1
3 1 2
5 3 2
3 3 1
 

Sample Output
-1
2
1

每个灯看做一个点，能互相照亮的点连边。从0、1、2三个源点求最短路，然后枚举这3个点到每个点的最短距离，得到答案。

#include"stdio.h"
#include"string.h"
#include"queue"
#include"vector"
#include"algorithm"
using namespace std;
#define N 205
const int inf=1000000;
int min(int a,int b)
{
    return a<b?a:b;
}
struct node
{
    int x,y,r;
}p[N];
int g[N][N];
int d[3][N];
int judge(int i,int j)   //判断两个灯是否相交
{
    int x,y,d,r;
    x=p[i].x-p[j].x;
    y=p[i].y-p[j].y;
    d=x*x+y*y;
    r=p[i].r+p[j].r;
    if(r*r>=d)
        return 1;
    return 0;
}
void spfa(int s,int n,int *dis)
{
    int i,mark[N];
    memset(mark,0,sizeof(mark));
    for(i=0;i<n;i++)
        dis[i]=inf;
    dis[s]=0;
    queue<int>q;
    q.push(s);
    mark[s]=1;
    while(!q.empty())
    {
        s=q.front();
        q.pop();
        mark[s]=0;
        for(i=0;i<n;i++)
        {
            if(dis[i]>dis[s]+g[s][i])
            {
                dis[i]=dis[s]+g[s][i];
                if(!mark[i])
                {
                    mark[i]=1;
                    q.push(i);
                }
            }
        }
    }
}
int main()
{
    int T,i,j,x,y,r,n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d%d%d",&x,&y,&r);
            p[i].x=x;p[i].y=y;p[i].r=r;
        }
        for(i=0;i<n;i++)
        {
            g[i][i]=0;
            for(j=0;j<i;j++)
            {
                if(judge(i,j))
                    g[i][j]=g[j][i]=1;
                else
                    g[i][j]=g[j][i]=inf;
            }
        }
        spfa(0,n,d[0]);
        spfa(1,n,d[1]);
        spfa(2,n,d[2]);
        int ans=inf;
        for(i=0;i<n;i++)
        {
            ans=min(ans,d[0][i]+d[1][i]+d[2][i]);
        }
        if(ans<inf)
            printf("%d\n",n-ans-1);
        else
            printf("-1\n");
    }
    return 0;
}


联系
我们