CSU 1116: Kingdoms(狀態壓縮+DP)

Kingdoms

Time Limit: 3 Sec  Memory Limit: 64 MB
Submit: 410  Solved: 127
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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

題意：有m條可修復的路，每個城市有相應的人口數，修復每條路需要指定的話費，問你在話費不超過K，最多能與多少人相通（可以當作能獲取多少資源）

解題思路：比賽時，由於到最後一個小時才接到這題，思路很混亂就開始上去寫了，比賽時將代碼寫戳了。言歸正傳：

    我們可以用二進制狀態表示訪問的每種情況，比如1010，表示城市二和四被訪問了，然後我們只要建立相應的狀態轉移方程即可。

<span style="font-size:18px;">#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cmath>
#include <vector>
#define INF 0x3f3f3f3f
using namespace std;
const int N=17;
int map[N][N];
int value[N];
int dp[1<<N];
int main()
{
    int n,m,l;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d %d",&n,&m,&l);
        int x,y,dis;
        memset(map,0x3f,sizeof(map));
        for(int i=1;i<=n;i++) scanf("%d",&value[i]);
        for(int i=0;i<m;i++){
                scanf("%d %d %d",&x,&y,&dis);
                map[y][x]=map[x][y]=min(map[x][y],dis);
        }
        memset(dp,0x3f,sizeof(dp));
        dp[1]=0;
        for(int i=2;i<(1<<n);i++) ///狀態
        {
            if(i&1){
                for(int j=1;j<=n;j++){
                    if(i &(1 <<(j-1))){
                        for(int k=1;k<=n;k++){
                            if(i &(1 <<(k-1))){
                                dp[i]=min(dp[i],dp[i &(~(1 <<(k-1)))]+map[k][j]);
                            }
                        }
                    }
                }
            }
        }
        int ans=0;
        for(int i=1;i<(1<<n);i++)
        {
            int sum=0;
            if(dp[i]<=l){
                for(int j=1;j<=n;j++){
                    if(i &(1 <<(j-1)))
                        sum+=value[j];
                }
            }
            ans=max(ans,sum);
        }
        printf("%d\n",ans);
    }
    return 0;
}

</span>