HDU 5934 Bomb 【图论缩点】（2016年中国大学生程序设计竞赛（杭州））

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10    Accepted Submission(s): 3


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）
 

 

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题目链接：

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

题目大意：

　　N个炸弹，坐标(x,y)，爆炸半径r，人为引爆的代价为c。

　　如果一个炸弹在已经爆炸的炸弹的爆炸半径内(上)，那么这个炸弹也会被引爆（连锁反应，不需要代价）

　　问至少多少代价可以把所有炸弹引爆。

题目思路：

　　【图论缩点】

　　如果A能引爆B那么A向B连一条边。

　　建完图之后用Tarjin或者Kosaraju强连通缩点。

　　缩完点之后的图是有向无环图。

　　再根据新的点连边，那些入度为0的点即为需要人为引爆的。统计代价即可。

 

 

  1 //
  2 //by coolxxx
  3 //#include<bits/stdc++.h>
  4 #include<iostream>
  5 #include<algorithm>
  6 #include<string>
  7 #include<iomanip>
  8 #include<map>
  9 #include<stack>
 10 #include<queue>
 11 #include<set>
 12 #include<bitset>
 13 #include<memory.h>
 14 #include<time.h>
 15 #include<stdio.h>
 16 #include<stdlib.h>
 17 #include<string.h>
 18 //#include<stdbool.h>
 19 #include<math.h>
 20 #pragma comment(linker,"/STACK:1024000000,1024000000")
 21 #define min(a,b) ((a)<(b)?(a):(b))
 22 #define max(a,b) ((a)>(b)?(a):(b))
 23 #define abs(a) ((a)>0?(a):(-(a)))
 24 #define lowbit(a) (a&(-a))
 25 #define sqr(a) ((a)*(a))
 26 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
 27 #define mem(a,b) memset(a,b,sizeof(a))
 28 #define eps (1e-8)
 29 #define J 10000
 30 #define mod 1000000007
 31 #define MAX 0x7f7f7f7f
 32 #define PI 3.14159265358979323
 33 #define N 1004
 34 #define M 1000004
 35 using namespace std;
 36 typedef long long LL;
 37 double anss;
 38 LL aans;
 39 int cas,cass;
 40 int n,m,lll,ans;
 41 int ll,sz;
 42 int fa[N],in[N],last[N],last1[N],num[N],cc[N];
 43 bool mark[N];
 44 struct Point
 45 {
 46     LL x,y,r;
 47     int c;
 48 }p[N];
 49 struct xxx
 50 {
 51     int next,to;
 52 }a[M],b[M];
 53 void add(int x,int y)
 54 {
 55     a[++lll].next=last[x];
 56     a[lll].to=y;
 57     last[x]=lll;
 58 }
 59 void link(int x,int y)
 60 {
 61     b[++ll].next=last1[x];
 62     b[ll].to=y;
 63     last1[x]=ll;
 64 }
 65 void dfs1(int now)
 66 {
 67     int i;
 68     mark[now]=1;
 69     for(i=last[now];i;i=a[i].next)
 70         if(!mark[a[i].to])
 71             dfs1(a[i].to);
 72     num[++sz]=now;
 73 }
 74 void dfs2(int now)
 75 {
 76     int i;
 77     mark[now]=1;
 78     fa[now]=m;
 79     cc[m]=min(cc[m],p[now].c);
 80     for(i=last1[now];i;i=b[i].next)
 81     {
 82         if(!mark[b[i].to])
 83             dfs2(b[i].to);
 84     }
 85 }
 86 void kosaraju()
 87 {
 88     int i;
 89     mem(mark,0);sz=0;m=0;
 90     for(i=1;i<=n;i++)
 91         if(!mark[i])
 92             dfs1(i);
 93     mem(mark,0);
 94     for(i=n;i;i--)
 95     {
 96         if(!mark[num[i]])
 97         {
 98             cc[++m]=MAX;
 99             dfs2(num[i]);
100         }
101     }
102 }
103 int main()
104 {
105     #ifndef ONLINE_JUDGEW
106 //    freopen("1.txt","r",stdin);
107 //    freopen("2.txt","w",stdout);
108     #endif
109     int i,j,k;
110     int x,y,z;
111 //    init();
112 //    for(scanf("%d",&cass);cass;cass--)
113     for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
114 //    while(~scanf("%s",s))
115 //    while(~scanf("%d%d",&n,&m))
116     {
117         printf("Case #%d: ",cass);
118         lll=0;ll=0;ans=0;
119         mem(in,0);mem(last,0);mem(last1,0);
120         scanf("%d",&n);
121         for(i=1;i<=n;i++)scanf("%lld%lld%lld%lld",&p[i].x,&p[i].y,&p[i].r,&p[i].c);
122         for(i=1;i<=n;i++)
123         {
124             for(j=1;j<=n;j++)
125             {
126                 if(i==j)continue;
127                 if(sqr(p[i].r)>=sqr(p[i].x-p[j].x)+sqr(p[i].y-p[j].y))
128                     add(i,j),link(j,i);
129             }
130         }
131         kosaraju();
132         for(i=1;i<=n;i++)
133         {
134             for(j=last[i];j;j=a[j].next)
135             {
136                 if(fa[a[j].to]==fa[i])continue;
137                 in[fa[a[j].to]]++;
138             }
139         }
140         for(i=1;i<=m;i++)
141             if(!in[i])
142                 ans+=cc[i];
143         printf("%d\n",ans);
144     }
145     return 0;
146 }
147 /*
148 //
149 
150 //
151 */
View Code

 

posted @ 2016-10-29 17:31  Cool639zhu  阅读(783)  评论(0)  编辑  收藏  举报