HDU5394 Bomb

题目：http://acm.hdu.edu.cn/showproblem.php?pid=5934

There are NN bombs needing exploding. 
Each bomb has three attributes: exploding radius riri, position (xi,yi)(xi,yi) and lighting-cost cici which means you need to pay cici 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.

InputFirst line contains an integer TT, which indicates the number of test cases.

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

In the following NN lines, the ith line contains four intergers xixi, yiyi, riri and cici, indicating the coordinate of ith bomb is (xi,yi)(xi,yi), exploding radius is riri and lighting-cost is cici.

Limits 
- 1≤T≤201≤T≤20 
- 1≤N≤10001≤N≤1000 
- −108≤xi,yi,ri≤108−108≤xi,yi,ri≤108 
- 1≤ci≤1041≤ci≤104OutputFor 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

1
5
0 0 1 5
1 1 1 6
0 1 1 7
3 0 2 10
5 0 1 4

Sample Output

Case #1: 15

题意：有n个炸弹，每个炸弹放在(x, y)这个位置，它能炸的范围是以 r 为半径的圆，手动引爆这颗炸弹所需代价是c，当一个炸弹爆炸时，

在它爆炸范围内的所有炸弹都将被它引爆，让求把所有的炸弹都引爆，所需的最少代价是多少？

建立单向图，然后缩点，每个点的权值都为它所在联通块的权值的小的那个，然后找到所有入度为0的点，将他们的权值加起来就是结果；

---

 #include<bits/stdc++.h>
 using namespace std;
 #define met(a, b) memset(a, b, sizeof(a))
 typedef long long LL;
 const int N=;
 const int INF=0x3f3f3f3f;
 const double eps=1e-;

 struct node{
     LL x,y,r;
 } a[N];

 int n,w[N],Min[N],dfn[N],low[N],vis[N];
 int Belong[N],Blocks,Stack[N],Top,Time,ind[N];
 vector<int> G[N];

 void Init()
 {
     for(int i=; i<=n; i++) G[i].clear();
     met(Min, INF); met(ind, );
     met(dfn, ); met(low, );
     met(Stack, ); met(vis, );
     met(Belong, );
     Blocks =Top=Time = ;
 }

 void Tajarn(int u)
 {
     low[u] = dfn[u] = ++Time;
     Stack[Top++] = u;
     vis[u] = ;
     int v;
     for(int i=;i<G[u].size();i++)
     {
         v=G[u][i];
         if(!dfn[v])
         {
             Tajarn(v);
             low[u]=min(low[u],low[v]);
         }
         else if(vis[v]) low[u]=min(low[u],dfn[v]);
     }

     if(low[u]==dfn[u])
     {
         ++Blocks;
         do
         {
             v = Stack[--Top];
             Belong[v] = Blocks;
             vis[v] = ;
         }while(u!=v);
     }
 }

 int main()
 {
     int T, t = ;
     scanf("%d", &T);
     while(T --)
     {
         scanf("%d", &n);
         Init();
         for(int i=;i<=n;i++) scanf("%I64d%I64d%I64d%d",&a[i].x,&a[i].y,&a[i].r,&w[i]);
         for(int i=; i<=n; i++)
         {
             for(int j=; j<=n; j++)
             {
                 LL d = (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*a[i].r>=d) G[i].push_back(j);
             }
         }

         for(int i=;i<=n;i++) if(!dfn[i]) Tajarn(i);
         for(int i=;i<=n;i++)
         {
             for(int j=;j<G[i].size();j++)
             {
                 int x = G[i][j];
                 int u = Belong[i], v = Belong[x];
                 if(u != v) ind[v] ++;
                 Min[u] = min(Min[u], w[i]);
                 Min[v] = min(Min[v], w[x]);
             }
         }

         int ans = ;
         for(int i=;i<=Blocks;i++) if(ind[i]==) ans+=Min[i];

         printf("Case #%d: %d\n", t++, ans);
     }
     return ;
 }

2018-09-10 16:58:27