HDU 5934

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2118    Accepted Submission(s): 687

Problem Description

There are N bombs needing exploding.

Each bomb has three attributes: exploding radius ri, position (xi,yi) and lighting-cost ci which means you need to pay ci cost making it explode.

If a un-lighting bomb is in or on the border the exploding area of another exploding one, the un-lighting bomb also will explode.

Now you know the attributes of all bombs, please use the minimum cost to explode all bombs.

 

Input

First line contains an integer T, which indicates the number of test cases.

Every test case begins with an integers N, which indicates the numbers of bombs.

In the following N lines, the ith line contains four intergers xi, yi, ri and ci, indicating the coordinate of ith bomb is (xi,yi), exploding radius is ri and lighting-cost is ci.

Limits
- 1≤T≤20
- 1≤N≤1000
- −108≤xi,yi,ri≤108
- 1≤ci≤104

 

Output

For every test case, you should output 'Case #x: y', where x indicates the case number and counts from 1 and y is the minimum cost.

 

Sample Input

150 0 1 51 1 1 60 1 1 73 0 2 105 0 1 4

 

Sample Output

Case #1: 15

 

Source

2016年中国大学生程序设计竞赛（杭州） 

 

题意：

一个炸弹有坐标，引爆半径，引爆代价，一个炸弹爆了，如果半径里有另一个炸弹，也会爆炸。

问你要把所有炸弹引爆，代码最小是多少。

POINT：

先建图，A半径有B A-B，B半径有A，B-A，所以是单向的。

然后tarjan缩点，每一个强连通分量都求出其中最小代价的炸弹。

之后取入度为0的炸弹引爆，他们的代价总和即是最小的。

注意如果把所以边存下来，开两百万的数组会mle。

就在遍历一次就好了，不会tle。

#include <iostream>#include <stdio.h>#include <vector>#include <math.h>#include <string.h>#include <stack>using namespace std;#define LL long longconst LL maxn = 2222;const LL inf = 0x3f3f3f3f;vector<LL>G[maxn];LL n;struct node{    LL x,y,r,c;} a[maxn];LL DFN[maxn],low[maxn],inq[maxn],cor[maxn],Min[maxn];LL cornum=0,tm=0;stack<LL>q;void init(){    for(LL i=1; i<=n; i++) G[i].clear();    memset(DFN,0,sizeof DFN);    memset(low,0,sizeof low);    memset(inq,0,sizeof inq);    memset(cor,0,sizeof cor);  //  memset(Min,inf,sizeof Min);    cornum=tm=0;    while(!q.empty())    {        q.pop();    }}void tarjan(LL u){    DFN[u]=low[u]=++tm;    inq[u]=1;    q.push(u);    for(LL i=0; i<G[u].size(); i++)    {        LL v=G[u][i];        if(DFN[v]==0)        {            tarjan(v);            if(low[u]>low[v]) low[u]=low[v];        }        else if(inq[v]&&low[u]>DFN[v])        {            low[u]=DFN[v];        }    }    if(DFN[u]==low[u])    {        cornum++;        LL v;        Min[cornum]=inf;        do        {            v=q.top();            Min[cornum]=min(Min[cornum],a[v].c);            cor[v]=cornum;            inq[v]=0;            q.pop();        }        while(v!=u);    }}int main(){    LL T;    scanf("%lld",&T);    LL p=0;    while(T--)    {        scanf("%lld",&n);        init();        for(LL i=1; i<=n; i++)        {            scanf("%lld %lld %lld %lld",&a[i].x,&a[i].y,&a[i].r,&a[i].c);        }        for(LL i=1; i<=n; i++)        {            for(LL j=1; j<=n; j++)            {                if(i!=j)                {                    double l=sqrt(1.0*(a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y));                    if(a[i].r>=l)                    {                        G[i].push_back(j);                    }                    if(a[j].r>=l)                    {                        G[j].push_back(i);                    }                }            }        }        for(LL i=1; i<=n; i++)        {            if(DFN[i]==0)            {                tarjan(i);            }        }        LL du[maxn];        memset(du,0,sizeof du);        for(LL i=1; i<=n; i++)        {            for(LL j=1; j<=n; j++)            {                if(i!=j)                {                    double l=sqrt(1.0*(a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y));                    if(cor[i]!=cor[j])                    {                        if(a[i].r>=l)                        {                            du[cor[j]]++;                        }                        if(a[j].r>=l)                        {                            du[cor[i]]++;                        }                    }                }            }        }        LL ans=0;        for(LL i=1;i<=cornum;i++){            if(du[i]==0){                ans+=Min[i];            }        }        printf("Case #%lld: ",++p);        printf("%lld\n",ans);    }    return 0;}