- 题目大意

某地发洪水，导致某些城市被淹而消失，现在想把剩下的零散的城市通过修路连接起来，已知现在有部分城市是连通的。可选择修的路有m条，城市总共有n个，给出了m条路的起点终点和修路花费，问最少可花多少钱能保证所有的城市连通。

- 解题思路

可以用kruskal。把边权排序然后并查集添加边即可。

- 代码

#include<algorithm>
#include<cstdio>
using namespace std;
int fa[505];struct Edge {int u, v, w;bool operator<(const Edge &rhs)const {return w < rhs.w;}
}e[25005];
void init(int n)
{for (int i = 1; i <= n; i++)fa[i] = i;
}int find(int x)
{if (x == fa[x])return x;else{return fa[x] = find(fa[x]);}
}bool Union(int x, int y)
{int fx = find(x), fy = find(y);if (fx == fy)return false;else{fa[fx] = fy;return true;}
}void kruskal(int m,int k)
{int sum = 0;sort(e, e + m);for (int i = 0; i < m; i++){int u = e[i].u, v = e[i].v, w = e[i].w;if (Union(u, v)){sum += w;k--;}}if (!k)printf("%d\n", sum);else{printf("-1\n");}
}
int main()
{int n,m,t,p,q,c,k,tmp;scanf_s("%d",&t);while (t--){scanf_s("%d%d%d",&n,&m,&k);init(n);for (int i = 0; i < m; i++)scanf_s("%d%d%d",&e[i].u,&e[i].v,&e[i].w);tmp = n-1;for (int j = 0; j < k; j++){scanf_s("%d%d",&p,&q);while (--p){scanf_s("%d", &c);if (Union(q,c)){tmp--;}}}kruskal(m, tmp);}return 0;
}

转载于:https://www.cnblogs.com/alpacadh/p/8449434.html

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D - Connect the Cities (HDU - 3371)

- 题目大意

- 解题思路

- 代码

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