洛谷p2330繁忙的都市题解

题面

根据题意来分析， 要求出你选了几条路， 最小生成树是能解的， 那么就直接输出n - 1条路即可， 至于最大值则走一遍最小生成树求出即可

这里提供最小生成树的两种方法

1. 克鲁斯卡尔

克鲁斯卡尔算法还是比较好理解的思路也简单， 就是排序然后按顺序来遍历， 找一个就把他放到并查集中。

#include <iostream>
#include <cstdio>
#include <algorithm>
#define MAXN 50005
using namespace std;
int n, m, tot, fa[MAXN];
struct node {
    int x, y, z;
}kls[MAXN];
bool cmp (node x, node y) {
    return x.z < y.z;
}
int find (int x) {
    if (fa[x] == x) return x;
    else return fa[x] = find (fa[x]);
}
int main () {
    scanf ("%d%d", &n, &m);
    for (int i = 1; i <= MAXN; i++)
        fa[i] = i;
    for (int i = 1; i <= m; i++) 
        scanf ("%d%d%d", &kls[i].x, &kls[i].y, &kls[i].z);
    sort (kls + 1, kls + m + 1, cmp);
    int ans;
    for (int i = 1; i <= m; i++) {
        int xx = find (kls[i].x), yy = find (kls[i].y);
        if (xx != yy) {
            fa[xx] = yy;
            tot++;
            ans = kls[i].z;
            if (tot == n - 1) break;
        }     
    }
    printf ("%d %d
", tot, ans);
    return 0;
}

2.prim算法

 有点类似于迪杰斯特拉。

#include <iostream>
#include <cstdio>
#include <cstring>
#define MAXN 5005
using namespace std;
int n, m, dis[MAXN], map[MAXN][MAXN]; 
bool vis[MAXN];
void prim () {
    dis[1] = 0;
    for (int i = 1; i <= n; i++) {
        int k = 0;
        for (int j = 1; j <= n; j++) 
            if (!vis[j] && dis[j] < dis[k])
                k = j;
        vis[k] = 1;
        for (int j = 1; j <= n; j++)
            if (!vis[j] && dis[j] > map[k][j])
                dis[j] = map[k][j];
    }
}
int main () {
    memset (map, 0x7f7f7f, sizeof (map));
    memset (dis, 0x7f7f7f, sizeof (dis));
    scanf ("%d%d", &n, &m);
    for (int i = 1; i <= m; i++) {
        int x, y, z;
        scanf ("%d%d%d", &x, &y, &z);
        if (map[x][y] > z)
            map[x][y] = map[y][x] = z;
    }
    prim();
    int big = 0;
    for (int i = 1; i <= n; i++)
        if (dis[i] > big)
            big = dis[i];
    printf ("%d %d
",n - 1, big);
    return 0;
}