51nod 1640 天气晴朗的魔法 最小生成树

题目链接:

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1640

题解:

先求最小生成树,记录最大边。

然后求最大生成树,保证所有边小于等于最小生成树的最大边。

#pragma comment(linker, "/STACK:102400000,102400000") 
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;

typedef long long LL;
const int maxn = 2e5 + 10;

int n, m;
struct Edge {
    int u, v, w;
}egs[maxn];

bool cmp1(const Edge& e1, const Edge& e2) {
    return e1.w < e2.w;
}

bool cmp2(const Edge& e1, const Edge& e2) {
    return e1.w > e2.w;
}

int fat[maxn];
int Find(int x) {
    return fat[x] = fat[x] == x ? x : Find(fat[x]);
}

LL kruskal(int &ma,int sign) {
    for (int i = 0; i <= n; i++) fat[i] = i;
    LL ret = 0;
    for (int i = 0; i < m; i++) {
        int pu = Find(egs[i].u);
        int pv = Find(egs[i].v);
        if (pu != pv) {
            if (sign) {
                if (egs[i].w <= ma) {
                    ret += egs[i].w;
                    fat[pv] = pu;
                }
            }
            else {
                ma = max(ma, egs[i].w);
                fat[pv] = pu;
            }
        }
    }
    return ret;
}

int main() {
    while (scanf("%d%d", &n, &m) == 2 && n) {
        int ma = -1;
        for (int i = 0; i < m; i++) {
            scanf("%d%d%d", &egs[i].u, &egs[i].v, &egs[i].w);
        }
        sort(egs, egs + m, cmp1);
        kruskal(ma,0);
        sort(egs, egs + m, cmp2);
        LL ans = kruskal(ma,1);
        printf("%lld
", ans);
    }
    return 0;
}
原文地址:https://www.cnblogs.com/fenice/p/5542319.html