CSU 1116 Kingdoms

1116: Kingdoms

Time Limit: 3 Sec  Memory Limit: 64 MB
Submit: 339  Solved: 103
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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

24 6 6500 400 300 2001 2 41 3 31 4 24 3 52 4 63 2 74 6 5500 400 300 2001 2 41 3 31 4 24 3 52 4 63 2 7

Sample Output

11001000

     这道题乍一看么什么思路，但是一看到点只有16个！所以果断枚举点！！因为想从边下手，然而边最多的时候有100条，所以如果每个点储存一个边的链表再去广搜遍历的时候就会有好多重复的情况，复杂度差不多是斐波那契的复杂度，所以果断放弃。

那现在谈谈怎么枚举点，很容易想到用01串来枚举，最多有16个点，所以就可以从0枚举得到1<<16，数量级都不到10的6次方，所以肯定行的，对每种情况进行一次最小生成树处理，最坏是16的平方，所以总的复杂度使程序在1秒内跑过是没问题的。那废话不多说，直接贴出代码，相应解释有注释说明：

#include <iostream>#include <algorithm>#include <iterator>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>using namespace std;#define INF 0x3f3f3f3fint map[20][20],people[20],low[20];int n,m,p;int vis[20];int prim(int t,int& sum){    memset(vis,0,sizeof(vis));    int pos,price,tot;    pos=1; price=0; tot=0;    for(int i=1;i<=n;i++)//寻找每种情况包含的点    {        vis[i]=t&1;        t=(t>>1);        if(vis[i])  tot++;//记录有多少点    }    if(vis[1]==0)   tot++;    for(int i=1;i<=n;i++)    {        if(vis[i])   low[i]=map[pos][i];        else    low[i]=INF;    }    low[1]=0; vis[1]=2; tot--;//只需要将tot个点进行最小生成树就行了    int MIN;    for(int k=1;k<=tot;k++)    {        MIN=INF; pos=1;        for(int i=1;i<=n;i++)        {            if(vis[i]==1&&MIN>low[i])            {                MIN=low[i]; pos=i;            }        }        if(pos==1)  return INF;//如果存在点扩展不到，就果断放弃这种情况        price+=low[pos]; sum+=people[pos]; vis[pos]=2;        for(int i=1;i<=n;i++)        {            if(vis[i]==1&&map[pos][i]<low[i])                low[i]=map[pos][i];        }    }    return price;}int main(){//    freopen("s","r",stdin);    int T;    scanf("%d",&T);    while(T--)    {        memset(map,0x3f,sizeof(map));        scanf("%d%d%d",&n,&m,&p);        for(int i=1;i<=n;i++)            scanf("%d",&people[i]);//记录每个城市的人数        for(int j=1;j<=m;j++)        {            int u,v,c;            scanf("%d%d%d",&u,&v,&c);            if(c<map[u][v])//处理重边，保留花费最少的边            {                map[u][v]=map[v][u]=c;            }        }        int price,sum,res=0;//        cout<<(1<<(n-1))<<endl;        for(int i=0;i<=(1<<(n));i++)//枚举点，对每个情况找一次最小生成树        {            sum=people[1];            price=prim(i,sum);//返回总花费            if(price<=p)    res=max(res,sum);//更新最大的人数        }        printf("%d\n",res);    }    return 0;}