HDU - 1863 畅通工程

来源：互联网发布：软件安装包制作编辑：程序博客网时间：2024/05/07 15:09

解题思路：并查集。根据造价排序，价格少的优先，如果不在同一个集合就合并，加上价格。最后判断有几个集合，若大于 1 个就说明所给方案不能连接所有村庄，输出 ?

#include <iostream>#include <algorithm>using namespace std;struct node {    int a, b, v;};node p[1010];int fa[1010];int gf(int x) {    if (fa[x] == x) return x;    fa[x] = gf(fa[x]);}bool judge(int x, int y) {    int fx = gf(x);    int fy = gf(y);    if (fx == fy) return 1;    fa[fx] = fy;    return 0;}bool cmp (node a, node b) {    return a.v < b.v;}int main() {    int N, M;    while (scanf("%d%d", &N, &M) && N) {        for (int i = 0; i < N; i++)            scanf("%d%d%d", &p[i].a, &p[i].b, &p[i].v);        for (int i = 0; i <= M; i++) fa[i] = i;        sort (p, p+N, cmp);        int sum = 0;        for (int i = 0; i < N; i++) {            if (!judge(p[i].a, p[i].b))                sum += p[i].v;        }        int flag = 0;        for (int i = 1; i <= M; i++)            if (fa[i] == i) flag++;        if (flag == 1) cout << sum << endl;        else cout << "?" << endl;    }return 0;}

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