湖南省第八届程序设计竞赛F

Problem F. Kingdoms

 

Akingdom has n cities numbered 1 to n, and some bidirectional roads connectingcities. The capital is always city 1.

 

After awar, all the roads of the kingdom are destroyed. The king wants to rebuild someof the roads to connect the cities, but unfortunately, the kingdom is runningout of money. The total cost of rebuilding roads should not exceed K.

 

Giventhe list of m roads that can be rebuilt (other roads are severelydamaged and cannot be rebuilt), the king decided to maximize the total populationin the capital and all other cities that are connected (directly or indirectly)with the capital (we call it "accessible population"), can you helphim?

 

Input

The first line of input contains a single integer T (T<=20), the numberof 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 npositive integers p_i(1<=p_i<=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 roadconnecting city u and v, whose rebuildingcost is c. Note that two cities canbe directly connected by more than one road, but a road cannot directlyconnect a city and itself.

 

Output

For each test case, print the maximalaccessible population.

 

Sample Input                           Output for 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

1100 1000

 

代码：

// Rujia Liu
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxn = 16 + 5;
const int maxm = 100 + 5;
const int INF = 1000000000;
int n, m;

int pa[maxn];
int findset(int x) { return pa[x] != x ? pa[x] = findset(pa[x]) : x; } 

struct Edge {
  int u, v, c;
  bool operator < (const Edge& rhs) const {
    return c < rhs.c;
  }
};

Edge e[maxm];

int MST() {
  sort(e, e+m);
  for(int i = 0; i < n; i++) pa[i] = i;
  int cnt = 0;
  int ans = 0;
  for(int i = 0; i < m; i++) {
    int u = e[i].u, v = e[i].v;
    u = findset(u);
    v = findset(v);
    int c = e[i].c;
    if(u != v) {
      pa[u] = v;
      ans += c;
      if(++cnt == n-1) return ans;
    }
  }
  return INF;
}

int in, im, K;
Edge ie[maxm];
int p[maxn];
int id[maxn];

int main() {
  int T;
  scanf("%d", &T);
  while(T--) {
    scanf("%d%d%d", &in, &im, &K);
    int p[maxn];
    Edge ie[maxn*maxn];
    for(int i = 0; i < in; i++) scanf("%d", &p[i]);
    for(int i = 0; i < im; i++) {
      int u, v, c;
      scanf("%d%d%d", &u, &v, &c);
      ie[i] = (Edge){ u-1, v-1, c };
    }

    int ans = p[0];
    int bestn = 1;
    for(int S = 0; S < (1<<(in-1)); S++) {
      n = 1;
      m = 0;
      memset(id, -1, sizeof(id));
      id[0] = 0; // capital
      for(int i = 0; i < in-1; i++)
        if(S & (1<<i)) id[i+1] = n++;
      for(int i = 0; i < im; i++) {
        int u = id[ie[i].u];
        int v = id[ie[i].v];
        if(u >= 0 && v >= 0) e[m++] = (Edge) { u, v, ie[i].c };
      }
      if(MST() <= K) {
        int tot = p[0];
        for(int i = 0; i < in-1; i++)
          if(S & (1<<i)) tot += p[i+1];
        if(tot > ans) { bestn = n; }
        ans = max(ans, tot);        
      }
    }

    printf("%d\n", ans);
  }
  return 0;
}