HDU5934-Bomb

Bomb

                                                                           Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                                                      Total Submission(s): 1185    Accepted Submission(s): 401

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

150 0 1 51 1 1 60 1 1 73 0 2 105 0 1 4

 

Sample Output

Case #1: 15

 

Source

2016年中国大学生程序设计竞赛（杭州）

 

题意：给你n个炸弹的坐标，爆炸半径和引爆需要的花费，问引爆所有炸弹需要的最少花费

解题思路：可以先建图后进行强联通缩点，然后重新建图后求出每个点引爆的最小花费，将所有入度为0的点需要引爆的最小花费进行求和

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include<cmath>using namespace std;#define LL long longconst int INF=0x3f3f3f3f;const int N=1005;int n;struct Edge{    int v,nt;} edge[N*N];struct node{    int w;    LL x,y,r;} x[N];int s[N],cnt;int dfn[N],low[N],id[N],dep,w[N];bool vis[N],instack[N];int res,graph[N][N],in[N];stack<int>st;void AddEdge(int u,int v){    edge[cnt].v=v;    edge[cnt].nt=s[u];    s[u]=cnt++;}void tarjan(int u){    st.push(u);    instack[u]=true;    vis[u]=true;    dfn[u]=low[u]=++dep;    for(int i=s[u]; ~i; i=edge[i].nt)    {        int v=edge[i].v;        if(!vis[v])        {            tarjan(v);            low[u]=min(low[u],low[v]);        }        else if(instack[v])            low[u]=min(low[u],dfn[v]);    }    if(dfn[u]==low[u])    {        int t;        do        {            id[t=st.top()]=res;            st.pop();            instack[t]=false;        }        while(t!=u);        res++;    }}void solve(){    res=0,dep=0;    while(!st.empty()) st.pop();    memset(vis,0,sizeof vis);    memset(instack,0,sizeof instack);    for(int i=1; i<=n; i++)        if(!vis[i]) tarjan(i);    memset(graph,0,sizeof graph);    memset(in,0,sizeof in);    for(int i=1; i<=n; i++)        for(int j=s[i]; ~j; j=edge[j].nt)        {            int v=edge[j].v;            if(id[i]!=id[v]) graph[id[i]][id[v]]=1,in[id[v]]++;        }    memset(w,INF,sizeof w);    for(int i=1; i<=n; i++)        w[id[i]]=min(w[id[i]],x[i].w);    int ans=0;    for(int i=0; i<res; i++)        if(!in[i])ans+=w[i];    printf("%d\n",ans);}int main(){    int t,cas=0;    scanf("%d",&t);    while(t--)    {        memset(s,-1,sizeof s);        scanf("%d",&n);        cnt=1;        for(int i=1; i<=n; i++) scanf("%lld%lld%lld%d",&x[i].x,&x[i].y,&x[i].r,&x[i].w);        for(int i=1; i<=n; i++)            for(int j=1; j<=n; j++)            {                if(i==j) continue;                if((x[i].x-x[j].x)*(x[i].x-x[j].x)+(x[i].y-x[j].y)*(x[i].y-x[j].y)<=x[i].r*x[i].r)                    AddEdge(i,j);            }        printf("Case #%d: ",++cas);        solve();    }    return 0;}