ybtoj最小生成树例题2新的开始

Posted 2021-05-16 SSL_ZZL

新的开始

• • • Link
    • 解题思路
    • Code

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Link

传送门
题目

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解题思路

很经典的模板题
设一个超级起点，所有点建发电站，视为是所有点向超级起点连边
再做最小生成树就ok了

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Code

#include <algorithm> 
#include <iostream>
#include <cstdio>

using namespace std;

struct DT{
	int x, y, s;
}a[95000];
int n, s, num, xx, yy, ans, fa[500];

bool cmp(const DT&k, const DT&l) {
	return k.s < l.s;
}

int find(int x) {
	if (fa[x] == -1) return x;
	fa[x] = find(fa[x]);
	return fa[x];
}

int main() {
	scanf("%d", &n);
	for (int i = 0; i <= n; i++)
		fa[i] = -1;  //我常用的超级起点是0， 所以初值不能设0
	
	for (int i = 1; i <= n; i++) {
		scanf("%d", &s);
		a[++num].x = 0, a[num].y = i, a[num].s = s;  //建发电站-》每个点向0作边
	}
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++) {
			scanf("%d", &s);
			a[++num].x = j, a[num].y = i, a[num].s = s;
		}
	sort(a + 1, a + 1 + (n * n + n), cmp);  //以下是普通的最小生成树
	for (int i = 1; i <= (n * n + n); i++) {
		xx = find(a[i].x), yy = find(a[i].y);
		if (xx != yy) {
			fa[xx] = yy;
			ans += a[i].s;
		}
	}
	printf("%d", ans);
}