Is it a tree?

题目:

http://poj.org/problem?id=1308

大意:

给你一堆树枝,让你判断最后能不能构成一棵树。

思路:根据树的定义判断即可

树:

1.入度只能为1或者0(根结点)

2.根结点只有一个

3.出度可以有多个


代码如下:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
#define MAXN 10010
#pragma comment(linker, "/STACK:102400000,102400000")
int pre[10010], son[10010], visit[10010];

int unionsearch(int root) //查找根结点+路径压缩(递归容易破栈)
{
	int root2;
	root2 = root;
	while(root != pre[root])
		root = pre[root];
	while(root2 != root)
	{
		int temp = pre[root2];
		pre[root2] = root;
		root2 = temp;
	}
	return root;
}

bool cmp()
{
	int root = 0;
	for(int i = 0; i < MAXN; ++i) //累计根结点的子节点个数
		if(visit[i])
			++son[unionsearch(i)];
	for(int i = 0; i < MAXN; ++i) //累计根结点的个数
		if(son[i] > 1)
			root++;
	if(root > 1) //判断为树或者为森林
		return false;
	else 
		return true;
}

int main()
{
	//freopen("Input.txt", "r", stdin);
	int start, end, x, y, flag, T = 0;	
	while(true)
	{
		flag = 0;
		for(int i = 0; i < MAXN; ++i)
			pre[i] = i;
		memset(son, 0, sizeof(son));
		memset(visit, 0, sizeof(visit));
		while(scanf("%d%d", &start, &end) && start && end)
		{
			if(start == -1 && end == -1) return 0;
			x = unionsearch(start);
			y = unionsearch(end);
			visit[start] = 1; // 树枝标记,忘了。调试了N久。
			visit[end] = 1;
			if(x == y || end != y) //x == y为环,end == y,入度为2
				flag = 1;
			else
				pre[end] = start;
		}
		if(flag)
			printf("Case %d is not a tree.\n", ++T);
		else
		{
			if(cmp())
				printf("Case %d is a tree.\n", ++T);
			else
				printf("Case %d is not a tree.\n", ++T);
		}
	}
	return 0;
}                

改进后的程序(2011/10/25)

 
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
#define MAXN 10010
int pre[10010], visit[10010];

int unionsearch(int root) //查找根结点+路径压缩
{
	int root2;
	root2 = root;
	while(root != pre[root])
		root = pre[root];
	while(root2 != root)
	{
		int temp = pre[root2];
		pre[root2] = root;
		root2 = temp;
	}
	return root;
}

bool cmp()
{
	int root = 0;
	for(int i = 0; i < MAXN; ++i) //累计根结点的子节点个数
		if(visit[i] && pre[i] == i)
			root++;
	if(root > 1) //判断为树或者为森林
		return false;
	else 
		return true;
}

int main()
{
	//freopen("Input.txt", "r", stdin);
	int start, end, x, y, flag, T = 0;	
	while(true)
	{
		flag = 0;
		for(int i = 0; i < MAXN; ++i)
			pre[i] = i;
		memset(visit, 0, sizeof(visit));
		while(scanf("%d%d", &start, &end) && start && end)
		{
			if(start == -1 && end == -1) return 0;
			x = unionsearch(start);
			y = unionsearch(end);
			visit[start] = 1;
			visit[end] = 1;
			if(x == y || end != y)
				flag = 1;
			else
				pre[end] = start;
		}
		if(flag)
			printf("Case %d is not a tree.\n", ++T);
		else
		{
			if(cmp())
				printf("Case %d is a tree.\n", ++T);
			else
				printf("Case %d is not a tree.\n", ++T);
		}
	}
	return 0;
}                        



你可能感兴趣的:(linker)