思路：建一個有向圖，指向能引爆對象，把強連通分量縮成一點，只要點燃圖中入度爲0的點即可。因爲入度爲0沒人能引爆，不爲0可以由別人引爆。
思路很簡單，但是早上寫的一直錯，改了半天了，推倒重來才過了...
#include<cstdio>
#include<set>
#include<stack>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn = 1000+5;
const int INF = 0x3f3f3f3f;
int head[maxn],dfn[maxn],low[maxn],vis[maxn],in[maxn],scc[maxn],val[maxn];
int tot,scc_int,index;
stack<int> st;
struct Edge{
int u,v,next;
}e[maxn*maxn];
struct Node{
ll x,y;
ll r,c;
}p[maxn];
void addEdge(int u,int v){
e[tot].u = u;
e[tot].v = v;
e[tot].next = head[u];
head[u] = tot++;
}
void tarjan(int u){
vis[u] = 1;
dfn[u] = low[u] = ++index;
st.push(u);
for(int i = head[u];i != -1;i = e[i].next){
int v = e[i].v;
if(!dfn[v]){
tarjan(v);
low[u] = min(low[u],low[v]);
}
else if(vis[v]){
low[u] = min(low[u],dfn[v]);
}
}
if(dfn[u] == low[u]){
scc_int++;
ll MIN = INF;
int v;
do{
v = st.top();
st.pop();
vis[v] = 0;
scc[v] = scc_int;
MIN = min(MIN,p[v].c);
}while(v != u);
val[scc_int] = MIN;
}
}
void init(){
tot = scc_int = index = 0;
while(!st.empty()) st.pop();
memset(head,-1,sizeof(head));
memset(dfn,0,sizeof(dfn));
memset(vis,0,sizeof(vis));
memset(in,0,sizeof(in));
}
int main(){
int T,n,Case = 1;
scanf("%d",&T);
while(T--){
scanf("%d",&n);
init();
for(int i = 1;i <= n;i++)
scanf("%lld%lld%lld%lld",&p[i].x,&p[i].y,&p[i].r,&p[i].c);
for(int i = 1;i <= n;i++){
for(int j = i + 1;j <= n;j++){
ll dis = (p[i].x - p[j].x)*(p[i].x - p[j].x) + (p[i].y - p[j].y)*(p[i].y - p[j].y);
if(p[i].r*p[i].r >= dis){
addEdge(i,j);
}
if(p[j].r*p[j].r >= dis){
addEdge(j,i);
}
}
}
for(int i = 1;i <= n;i++){
if(!dfn[i])
tarjan(i);
}
for(int i = 1; i<= n;i++){
for(int j = head[i];~j;j = e[j].next){
int v = e[j].v;
if(scc[i] != scc[v]){
in[scc[v]]++;
}
}
}
ll ans = 0;
for(int i = 1;i <= scc_int;i++){
if(in[i] == 0){
ans += val[i];
}
}
printf("Case #%d: %lld\n",Case++,ans);
}
return 0;
}