csuoj 1116: Kingdoms

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1116

1116: Kingdoms

Time Limit: 3 Sec  Memory Limit: 64 MB
Submit: 293  Solved: 82
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Description

A kingdom has n cities numbered 1 to n, and some bidirectional roads connecting cities. The capital is always city 1.

After a war, all the roads of the kingdom are destroyed. The king wants to rebuild some of the roads to connect the cities, but unfortunately, the kingdom is running out of money. The total cost of rebuilding roads should not exceed K.

Given the list of m roads that can be rebuilt (other roads are severely damaged and cannot be rebuilt), the king decided to maximize the total population in the capital and all other cities that are connected (directly or indirectly) with the capital (we call it "accessible population"), can you help him?

Input

The first line of input contains a single integer T (T<=20), the number of test cases. 

Each test case begins with three integers n(4<=n<=16), m(1<=m<=100) and K(1<=K<=100,000). 

The second line contains n positive integers pi (1<=pi<=10,000), the population of each city. 

Each of the following m lines contains three positive integers u, v, c (1<=u,v<=n, 1<=c<=1000), representing a destroyed road connecting city u and v, whose rebuilding cost is c. 

Note that two cities can be directly connected by more than one road, but a road cannot directly connect a city and itself.

Output

For each test case, print the maximal accessible population.

Sample Input

2
4 6 6
500 400 300 200
1 2 4
1 3 3
1 4 2
4 3 5
2 4 6
3 2 7
4 6 5
500 400 300 200
1 2 4
1 3 3
1 4 2
4 3 5
2 4 6
3 2 7

Sample Output

1100
1000

HINT

 

Source

湖南省第八屆大學生計算機程序設計競賽

 

分析；

此題能夠先肯定1是在點集裏，而後暴力枚舉其它城市是否要連，對每一個枚舉的結果求最小生成樹，選出符合條件的最優解。

 

 

AC代碼:

 

 

  1 #include <iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 using namespace std;
  5 const int maxn=0x3f3f3f3f;
  6 int n,m,k;
  7 int population[20];
  8 int Map[20][20];
  9 bool exits[20];
 10 bool vis[20];
 11 int d[20];
 12 int p;
 13 void init(int n)
 14 {
 15     for(int i=1; i<=n; i++)
 16     {
 17         for(int j=1; j<=n; j++)
 18         {
 19             Map[i][j]=(i!=j?maxn:0);
 20         }
 21     }
 22 }
 23 int prim()
 24 {
 25     int sum=0;
 26     memset(vis,false,sizeof(vis));
 27     for(int i=1; i<=n; i++)
 28     {
 29         d[i]=Map[1][i];
 30     }
 31     vis[1]=true;
 32     for(int i=2; i<=n; i++)
 33     {
 34 
 35         int min=maxn,mini;
 36         for(int j=1; j<=n; j++)
 37         {
 38             if(!vis[j]&&exits[j]&&d[j]<min)
 39             {
 40                 min=d[j];
 41                 mini=j;
 42             }
 43         }
 44         if(min==maxn) break;
 45         sum+=min;
 46         vis[mini]=true;
 47         for(int k=1; k<=n; k++)
 48         {
 49             if(exits[k]&&!vis[k]&&Map[mini][k]<d[k])
 50             {
 51                 d[k]=Map[mini][k];
 52             }
 53         }
 54 
 55 
 56     }
 57     int cnt1=0,cnt2=0;
 58     for(int i=1;i<=n;i++){
 59         cnt1+=exits[i];
 60     }
 61     for(int i=1;i<=n;i++){
 62         cnt2+=vis[i];
 63     }
 64     if(cnt1==cnt2)
 65     return sum;
 66     else return maxn;
 67 
 68 }
 69 
 70 void dfs(int cur)
 71 {
 72     if(cur>n)
 73     {
 74         int pr=prim();
 75         int temp=0;
 76         if(pr<=k)
 77         {
 78             for(int i=1; i<=n; i++)
 79             {
 80                 if(exits[i])
 81                 {
 82                     temp+=population[i];
 83                 }
 84             }
 85             if(temp>p) p=temp;
 86         }
 87         return ;
 88     }
 89     for(int i=0; i<2; i++)
 90     {
 91         exits[cur]=i==1?true:false;
 92         dfs(cur+1);
 93     }
 94 }
 95 int main()
 96 {
 97     int t;
 98     scanf("%d",&t);
 99     while(t--)
100     {
101         scanf("%d%d%d",&n,&m,&k);
102         for(int i=1; i<=n; i++)
103         {
104             scanf("%d",&population[i]);
105         }
106         init(n);
107         for(int i=0; i<m; i++)
108         {
109             int from,to,c;
110             scanf("%d%d%d",&from,&to,&c);
111             if(Map[from][to]>c)
112             {
113                 Map[from][to]=Map[to][from]=c;
114             }
115         }
116         p=population[1];
117         exits[1]=true;
118         dfs(2);
119         printf("%d\n",p);
120     }
121     return 0;
122 }

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