uva-208-枚举-并查集

Posted 2020-10-10 菜菜

题意：

给你一个图,从1到指点的点有多少种不同的路径,用了并查集剪枝,如果当前节点的根不是指定的节点,直接返回,会超时

time limit了俩次,wa了俩次,PE俩次

#include <iostream>
#include <stdio.h>
#include<memory.h>
using namespace std;

#define null NULL
const int N = 25;
int final = -1;
int mindex[N];
int map[N][N];
int vis[N];
int res[N];
int n, m;
int total = 0;

int set[N];
int p(int s)
{
	return set[s] == s ? s : set[s] = p(set[s]);
}
void add(int s, int e)
{
	int p1 = p(s);
	int p2 = p(e);
	int t = s == final ? s : e;
	if (t != final)
		t = p1 == final ? p1 : p2;
	if (t == final)
	{
		set[t] = final;
		set[p1] = final;
		set[p2] = final;
		set[s] = final;
		set[e] = final;
		return;
	}
	if (p1 < p2)
	{
		set[p2] = p1;
	}
	else
	{
		set[p1] = p2;
	}
}

void read()
{
	for (int i = 0; i < N; i++)
		set[i] = i;

	int s, e;
	while (cin >> s >> e)
	{
		if (s == 0 && e == 0)
			return;
		map[s][mindex[s]++] = e;
		map[e][mindex[e]++] = s;
		add(s, e);
	}
}

void sort(int l, int t, int a[])
{
	for (int i = 0; i < t; i++)
	{
		for (int j = 1; j < t - i; j++)
		{
			if (a[j - 1] > a[j])
			{
				int t = a[j - 1];
				a[j - 1] = a[j];
				a[j] = t;
			}
		}
	}
}

void dfs(int cur, int rl)
{
	if (cur == final)
	{
		printf("%d",res[0]);
		for (int i = 1; i <= rl; i++)
		{
			printf(" %d", res[i]);
		}
		cout << endl;
		++total;
		return;
	}
	if (p(cur) != final)
		return;
	for (int i = 0; i < mindex[cur]; i++)
	{
		if (vis[map[cur][i]] == 0)
		{
			vis[map[cur][i]] = 1;
			res[rl + 1] = map[cur][i];
			dfs(map[cur][i], rl + 1);
			vis[map[cur][i]] = 0;
		}
	}
}

int main(int argc, char* argv[])
{
	freopen("C:\\Users\\zzzzz\\Desktop\\1.txt", "r", stdin);
	int t = 1;
	string str =
			"There are %d routes from the firestation to streetcorner %d.";
	while (cin >> final)
	{
		memset(map, 0, sizeof(map));
		memset(mindex, 0, sizeof(mindex));
		memset(vis, 0, sizeof(vis));
		read();
		total = 0;
		cout << "CASE " << t << ":" << endl;
		for (int i = 1; i <= final; i++)
			sort(i, mindex[i], map[i]);
		vis[1] = 1;
		res[0] = 1;
		dfs(1, 0);
		printf(str.c_str(), total, final);
		cout<<endl;
		++t;
	}

	return 0;
}