题目链接：​​Bomb​​

题目大意：现在有一些炸弹，有他的坐标，爆炸半径范围以及爆炸缩需要的价值，问要引爆所有的炸弹最少需要多少花费

题目思路：因为A能引爆B导致B引爆C，所以A能引爆C，但是A引爆B不代表B能引爆A，因为爆炸半径不一样嘛，所以建有向边，强连通分量缩点，然后找入度为零的联通快，必须引爆，不然没办法全部炸掉，引爆的花费是这个联通块的里面所有点的最小价值，然后加起来就好了

#include <map>
#include <set>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <utility>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 20010;
const int maxm = 1e5+10;

vector<int>Edge[maxm];
stack<int>S;

int T,n;
ll x[maxn],y[maxn],r[maxn],c[maxn];

int Dfn[maxn],Low[maxn],sccno[maxn],tclock,sccnt;
int InDeg[maxn],mincostscc[maxn];

void tarjan(int u){
    Dfn[u] = Low[u] = ++tclock;
    S.push(u);

    for(int i = 0;i < Edge[u].size();i++){
        int v = Edge[u][i];
        if(!Dfn[v]){
            tarjan(v);
            Low[u] = min(Low[u],Low[v]);
        }
        else if(!sccno[v]){
            Low[u] = min(Low[u],Dfn[v]);
        }
    }

    if(Low[u] == Dfn[u]){
        sccnt ++;
        int v = -1;
        while(v != u){
            v = S.top();
            S.pop();
            sccno[v] = sccnt;
            mincostscc[sccnt] = min(mincostscc[sccnt],(int)c[v]);
        }
    }
}

void findscc(int n){
    tclock = sccnt = 0;
    memset(Dfn,0,sizeof(Dfn));
    memset(Low,0,sizeof(Low));
    memset(sccno,0,sizeof(sccno));
    memset(mincostscc,0x3f3f3f3f,sizeof(mincostscc));
    for(int i = 1;i <= n;i++)
        if(!Dfn[i]) tarjan(i);
}

bool judge(ll x1,ll y1,ll x2,ll y2,ll r){
    if((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2) <= r*r) return true;
    return false;
}

int solve(int n){
    memset(InDeg,0,sizeof(InDeg));
    for(int u = 1;u <= n;u++){
        for(int i = 0;i < Edge[u].size();i++){
            int v = Edge[u][i];
            if(sccno[u] != sccno[v]){
                InDeg[sccno[v]]++;
            }
        }
    }
    int minn = 0;
    for(int i = 1;i <= sccnt;i++){
        if(InDeg[i] == 0) minn += mincostscc[i];
    }
    return minn;
}

int main(){
    scanf("%d",&T);
    for(int Case = 1;Case <= T;Case++){
        scanf("%d",&n);
        for(int i = 1;i <= n;i++) scanf("%lld%lld%lld%lld",&x[i],&y[i],&r[i],&c[i]);
        for(int i = 1;i <= n;i++) Edge[i].clear();
        for(int i = 1;i <= n;i++){
            for(int j = i+1;j <= n;j++){
                if(judge(x[i],y[i],x[j],y[j],r[i])) Edge[i].push_back(j);
                if(judge(x[i],y[i],x[j],y[j],r[j])) Edge[j].push_back(i);
            }
        }
        findscc(n);
        printf("Case #%d: %d\n",Case,solve(n));
    }
    return 0;
}