Bomb
Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4098 Accepted Submission(s): 1264
Problem Description
There are N
bombs needing exploding.
Each bomb has three attributes: exploding radius r
i![]()
, position (x
i
,y
i
)
and lighting-cost c
i![]()
which means you need to pay c
i![]()
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.
Each bomb has three attributes: exploding radius r
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 x
i![]()
, y
i![]()
, r
i![]()
and c
i![]()
, indicating the coordinate of ith bomb is (x
i
,y
i
)
, exploding radius is r
i![]()
and lighting-cost is c
i![]()
.
Limits
- 1≤T≤20
- 1≤N≤1000
- −10
8
≤x
i
,y
i
,r
i
≤10
8![]()
- 1≤c
i
≤10
4![]()
Every test case begins with an integers N
In the following N
Limits
- 1≤T≤20
- 1≤N≤1000
- −10
- 1≤c
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
#include<bits/stdc++.h> #define REP(i, a, b) for(int i = (a); i <= (b); ++ i) #define REP(j, a, b) for(int j = (a); j <= (b); ++ j) #define PER(i, a, b) for(int i = (a); i >= (b); -- i) using namespace std; const int maxn=1e5+5; template <class T> inline void rd(T &ret){ char c; ret = 0; while ((c = getchar()) < '0' || c > '9'); while (c >= '0' && c <= '9'){ ret = ret * 10 + (c - '0'), c = getchar(); } } struct node{ long long x,y,r,val; }p[maxn]; int dfn[maxn],low[maxn],in[maxn],T,n,tot,pos[maxn],cnt,vis[maxn],w[maxn]; vector<int>g[maxn]; stack<int>sk; void tarjan(int cur){ dfn[cur]=low[cur]=++tot; vis[cur]=1; sk.push(cur); for(int i=0;i<g[cur].size();i++){ int to=g[cur][i]; if(!dfn[to]){ tarjan(to); low[cur]=min(low[cur],low[to]); } else if(vis[to])low[cur]=min(low[cur],dfn[to]); } if(dfn[cur]==low[cur]){ cnt++; int now; do{ now=sk.top(); sk.pop(); pos[now]=cnt; vis[now]=0; w[cnt]=min(w[cnt],(int)p[now].val); }while(now!=cur); } } int main() { int many=1; rd(T); while(T--){ rd(n); memset(dfn,0,sizeof(dfn)); memset(low,0,sizeof(low)); memset(in,0,sizeof(in)); memset(w,127,sizeof(w)); memset(vis,0,sizeof(vis)); memset(pos,0,sizeof(pos)); tot=0,cnt=0; REP(i,1,n)g[i].clear(); REP(i,1,n){ scanf("%lld%lld%lld%lld",&p[i].x,&p[i].y,&p[i].r,&p[i].val); for(int j=1;j<i;j++){ if((p[i].x-p[j].x)*(p[i].x-p[j].x)+(p[i].y-p[j].y)*(p[i].y-p[j].y)<=p[i].r*p[i].r){ g[i].push_back(j); } if((p[i].x-p[j].x)*(p[i].x-p[j].x)+(p[i].y-p[j].y)*(p[i].y-p[j].y)<=p[j].r*p[j].r){ g[j].push_back(i); } } } REP(i,1,n)if(!dfn[i])tarjan(i); REP(i,1,n){ for(int j=0;j<g[i].size();j++){ int to=g[i][j]; if(pos[i]!=pos[to])in[pos[to]]++; } } long long ans=0; REP(i,1,cnt){ if(!in[i])ans+=w[i]; } cout<<"Case #"<<many++<<": "; cout<<ans<<endl; } return 0; }