POJ1251 Jungle Roads(kru)

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给出n - 1组边长, 构成一张图, 要求删除一些边使得边长和最小且任意两点可到达.

最小生成树, 套一个kru模板即可. 并查集初始化, 读入数据后对边长进行排序, 若并查集确定两点不在同一连通分量则将两点连接并存储

所得到的边.

AC代码:

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"using namespace std;const int INF = 0x3f3f3f3f;const int MAXN = 100;struct node{/* data */int u, v, w;}map[MAXN];int par[30], n;int find(int x){if(x == par[x]) return x;return par[x] = find(par[x]);}bool cmp(node a, node b){return a.w < b.w;}int main(int argc, char const *argv[]){while(cin >> n && n) {for(int i = 0; i < 30; ++i)par[i] = i;int num = 0, ans = 0;for(int i = 0; i < n - 1; ++i) {char c;int k;cin >> c >> k;for(int j = 0; j < k; ++j) {char ch;int x;cin >> ch >> x; map[num].u = c - 'A';map[num].v = ch - 'A';map[num++].w = x;}}sort(map, map + num, cmp);for(int i = 0; i < num; ++i) {int x = find(map[i].u), y = find(map[i].v);if(x != y) {ans += map[i].w;par[x] = y;}}cout << ans << endl;} return 0;}

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