Uva--1151(图论,MST,二进制枚举)

2014-09-09 16:39:37

题目链接

思路:做完这题,对最小生成树的理解也就差不多了。直接暴力可搞,但可以优化。先对原图进行一次Kruskal,求出最优n-1条边,然后用二进制枚举方法(循环 or DFS),考虑每个套餐的包含点(m个),买这个套餐相当于加进来m-1条权为0的边,然后再用一次Kruskal,先考虑m-1条0权边再考虑第一次kru出来的n-1条边即可。

(被这输出坑了两发 QAQ)

  1 /*************************************************************************
  2     > File Name: Uva1151.cpp
  3     > Author: Nature
  4     > Mail: 564374850@qq.com
  5     > Created Time: Tue 09 Sep 2014 02:27:23 PM CST
  6 ************************************************************************/
  7 
  8 #include <cstdio>
  9 #include <cstring>
 10 #include <cstdlib>
 11 #include <cmath>
 12 #include <vector>
 13 #include <queue>
 14 #include <iostream>
 15 #include <algorithm>
 16 using namespace std;
 17 typedef long long ll;
 18 const int INF = 0x3f3f3f3f;
 19 
 20 struct edge{
 21     int u,v;
 22     double dis;
 23 }e[1000005];
 24 
 25 edge te[10005];
 26 
 27 struct pac{
 28     int n,f;
 29     int v[1005];
 30 }p[10];
 31 
 32 int T,n,s,x[1005],y[1005],cnt;
 33 int fa[1005],ncnt,tcnt,ans;
 34 
 35 int Dis(int a,int b){
 36     return (x[b]-x[a])*(x[b]-x[a]) + (y[b]-y[a])*(y[b]-y[a]);
 37 }
 38 
 39 int Find(int x){ return fa[x] == x ? x : fa[x] = Find(fa[x]);}
 40 
 41 void Kruskal(){
 42     for(int i = 1; i <= n; ++i) fa[i] = i;
 43     ncnt = 0;
 44     for(int i = 1; i <= cnt; ++i){
 45         int x = Find(e[i].u);
 46         int y = Find(e[i].v);
 47         if(x != y){
 48             fa[y] = x;
 49             e[++ncnt] = e[i];
 50         }
 51     }
 52 }
 53 
 54 bool cmp(edge a,edge b){
 55     return a.dis < b.dis;
 56 }
 57 
 58 void Cal(int st){
 59     int x,y,fee = 0;
 60     tcnt = 0;
 61     for(int i = 1; i <= s; ++i){
 62         if(st & (1 << i)){
 63             for(int j = 1; j < p[i].n; ++j){
 64                 te[++tcnt].u = p[i].v[j];
 65                 te[tcnt].v = p[i].v[j + 1];
 66                 te[tcnt].dis = 0;
 67             }
 68             fee += p[i].f;
 69         }
 70     }
 71     for(int i = 1; i <= n; ++i) fa[i] = i;
 72     for(int i = 1; i <= tcnt; ++i){
 73         x = Find(te[i].u);
 74         y = Find(te[i].v);
 75         if(x != y) fa[y] = x;
 76     }
 77     for(int i = 1; i <= ncnt; ++i){
 78         x = Find(e[i].u);
 79         y = Find(e[i].v);
 80         if(x != y){
 81             fa[y] = x;
 82             fee += e[i].dis;
 83         }
 84     }
 85     if(fee < ans){
 86         ans = fee;
 87     }
 88 }
 89 
 90 void Dfs(int p,int st){
 91     if(p > s){
 92         Cal(st);
 93         return;
 94     }
 95     Dfs(p + 1,st | (1 << p));
 96     Dfs(p + 1,st);
 97 }
 98 
 99 int main(){
100     scanf("%d",&T);
101     while(T--){
102         scanf("%d%d",&n,&s);
103         for(int i = 1; i <= s; ++i){
104             scanf("%d%d",&p[i].n,&p[i].f);
105             for(int j = 1; j <= p[i].n; ++j)
106                 scanf("%d",&p[i].v[j]);
107         }
108         for(int i = 1; i <= n; ++i)
109             scanf("%d%d",x + i,y + i);
110         cnt = 0;
111         for(int i = 1; i <= n; ++i)
112             for(int j = i + 1; j <= n; ++j){
113                 if(i == j) continue;
114                 e[++cnt].u = i;
115                 e[cnt].v = j;
116                 e[cnt].dis = Dis(i,j);
117             }
118         sort(e + 1,e + cnt + 1,cmp);
119         Kruskal();
120         ans = INF;
121         Dfs(1,0);
122         printf("%d\n",ans);
123         if(T) printf("\n");
124     }
125     return 0;
126 }

 

posted @ 2014-09-09 16:43  Naturain  阅读(238)  评论(0编辑  收藏  举报