HDU5934 Bomb(2016杭州CCPC第二题)（强连通缩点）

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26    Accepted Submission(s): 10

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

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

 

Source

2016年中国大学生程序设计竞赛（杭州）

【总结】感觉自己智商就是一坨屎，当时在杭州比赛的时候，看到这道题，就想到了强连通分量，但我想的方向错了，我一直想着把一些互相可以到达的点缩成一个点或者找一棵树的根，却没有想到从缩点后点的入度下手，真是亏了我前段时间刷了好多强连通的题。哎 还是刷题太少，思路太狭窄而且比赛临场经验不足。真的很遗憾，我们当时是铁牌第一，我刚才照着模板敲了一下就过了，要是当时想到入度，就可以拿铜牌了。是该总结一下了，确实不能盲目刷题，还是要总结体型，拓展思路，不能再留下遗憾了。下面附上两个星期前就该A掉的代码。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
typedef long long ll;
using namespace std;
const int N = 1005;
const int M = 24005;
int vis[N],dfn[N],low[N],head[N],stack1[N],num[N],in[N];
ll cost[N];
int n,m,tot,son,maxn,tim,top,cut;
ll ans;
struct EDG{int to,next;}edg[N*N];
struct node{ll x,y,r,c;}a[N];
bool cmp(node f,node g){return f.c<g.c;}
void add(int u,int v){
    edg[tot].to=v;edg[tot].next=head[u];head[u]=tot++;
}
void init(){
    met(head,-1);
    tot=tim=top=cut=0;
    met(vis,0);
    met(edg,0);
    met(in,0);
    met(cost,inf);
    met(stack1,0);met(num,0);met(dfn,0);met(low,0);
}
void Tarjan(int u) {
    int v;
    low[u] = dfn[u] = ++tim;
    stack1[top++] = u;
    vis[u] = 1;
    for(int e = head[u]; e != -1; e = edg[e].next){
        v = edg[e].to;
        if(!dfn[v]){
            Tarjan(v);
            low[u] = min(low[u], low[v]);
        }else if(vis[v]){
            low[u] = min(low[u], dfn[v]);
        }
    }
    if(low[u] == dfn[u]){
        cut++;
        do{
            v = stack1[--top];
            num[v] = cut;
            cost[cut]=min(cost[cut],a[v].c);
            vis[v] = 0;
        }while(u != v);
    }
}
int main() {
    int T;
    scanf("%d",&T);
    for(int t=1;t<=T;t++){
        init();
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%lld%lld%lld%lld",&a[i].x,&a[i].y,&a[i].r,&a[i].c);
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                if(i==j)continue;
                if((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y)<=a[i].r*a[i].r){
                    add(i,j);
                }
            }
        }
        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!=-1; j=edg[j].next) {
                int v=edg[j].to;
                if(num[i]!=num[v])in[num[v]]++;
            }
        }
        ans=0;
        for(int i=1;i<=cut;i++){
            if(!in[i])ans+=cost[i];
        }
        printf("Case #%d: %lld
",t,ans);
    }
    return 0;
}