最小生成树的计数问题

时间:2019-12-11
本文章向大家介绍最小生成树的计数问题,主要包括最小生成树的计数问题使用实例、应用技巧、基本知识点总结和需要注意事项,具有一定的参考价值,需要的朋友可以参考一下。

https://www.luogu.com.cn/problemnew/solution/P4208

看这里的题解就够了其实

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#define maxn 1010
using namespace std;
const int mod = 31011;
int n, m;
struct Node {
	int be, en, len, cnt;
}que[maxn];

vector<Node>ins;

bool bml(Node a, Node b) {
	return a.len < b.len;
}
int par[111];
int find(int x) {
	if (par[x] == -1)  return x;
	return find(par[x]);
}

int ans = 0;

int dfs(int x, int i, int k) {
	if (i == ins[x].en + 1) {
		if (k == ins[x].cnt) ans++;
		return 0;
	}
	int a = find(que[i].be);
	int b = find(que[i].en);
	if (a != b) {
		par[a] = b;
		dfs(x, i + 1, k + 1);//选
		par[a] = -1;
		par[b] = -1;
	}
	dfs(x, i + 1, k);//不选
}
int main() {
	memset(par, -1, sizeof(par));
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; i++) {
		scanf("%d%d%d", &que[i].be, &que[i].en, &que[i].len);
	}
	sort(que + 1, que + m + 1, bml);
	que[m + 1].len = -1;
	int cnt = 0;
	int nn = 0;
	for (int i = 1; i <= m; i++) {
		int a = find(que[i].be);
		int b = find(que[i].en);
		if (a != b) {
			par[a] = b;
			cnt++;
			nn++;
		}
		if (que[i].len != que[i+1].len) {
			Node an;
			an.en = i;
			an.len = que[i].len;
			an.cnt = cnt;
			cnt = 0;
			if (ins.size() == 0) an.be = 1;
			else an.be = ins[ins.size() - 1].en + 1;
			ins.push_back(an);
		}
	}
	if (nn != n - 1) {
		printf("0\n");
		return 0;
	}
	memset(par, -1, sizeof(par));
	int cns = 1;

	for (int i = 0; i < ins.size(); i++) {
		dfs(i, ins[i].be, 0);
		cns = (cns*ans) % mod;
	
		ans = 0;
		for (int j = ins[i].be; j <= ins[i].en; j++) {
			int a = find(que[j].be);
			int b = find(que[j].en);
			if (a != b) {
				par[a] = b;
			}
		}
	}
	printf("%d\n", cns);
	return 0;
}

  

原文地址:https://www.cnblogs.com/lesning/p/12022107.html