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

[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

Case #1: 15

[align=left]Source[/align]
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
,dfn
,low
,head
,stack1
,num
,in
;
ll cost
;
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
;
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\n",t,ans);
}
return 0;
}