POJ 1308 && HDU 1325 Is It A Tree?（并查集）

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Is It A Tree?

Time Limit: 1000MSMemory Limit: 10000KTotal Submissions: 34386Accepted: 11650

Description

A tree is a well-known data structure that is either empty (null, void, nothing) or is a set of one or more nodes connected by directed edges between nodes satisfying the following properties.


There is exactly one node, called the root, to which no directed edges point.

Every node except the root has exactly one edge pointing to it.

There is a unique sequence of directed edges from the root to each node.

For example, consider the illustrations below, in which nodes are represented by circles and edges are represented by lines with arrowheads. The first two of these are trees, but the last is not.



In this problem you will be given several descriptions of collections of nodes connected by directed edges. For each of these you are to determine if the collection satisfies the definition of a tree or not.

Input

The input will consist of a sequence of descriptions (test cases) followed by a pair of negative integers. Each test case will consist of a sequence of edge descriptions followed by a pair of zeroes Each edge description will consist of a pair of integers; the first integer identifies the node from which the edge begins, and the second integer identifies the node to which the edge is directed. Node numbers will always be greater than zero.

Output

For each test case display the line “Case k is a tree.” or the line “Case k is not a tree.”, where k corresponds to the test case number (they are sequentially numbered starting with 1).

Sample Input

6 8  5 3  5 2  6 4 
5 6  0 0
8 1  7 3  6 2  8 9  7 5 
7 4  7 8  7 6  0 0
3 8  6 8  6 4 
5 3  5 6  5 2  0 0 
-1 -1

Sample Output

Case 1 is a tree. 
Case 2 is a tree. 
Case 3 is not a tree.

Source

North Central North America 1997

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思路

    这题就是给你一堆边，让你判断这是不是一颗树，我的方法非常奇怪，并不是用并查集做的。我是先判断是否有根，然后如果有节点的入度大于一，那就不是一颗树了。不过好像用并查集更简单…
    P.S.注意读入方法和输出格式。并且空树也算一颗树。

Code

非并查集做法

#pragma GCC optimize(3)#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>#include<cctype>#include<climits>#include<cstdlib>#include<cmath>#include<queue>#include<stack>#include<climits>#include<vector>using namespace std;typedef long long ll;inline void readInt(int &x) {    x=0;int f=1;char ch=getchar();    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();    x*=f;}inline void readLong(ll &x) {    x=0;int f=1;char ch=getchar();    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();    x*=f;}/*================Header Template==============*/int u,v,ru[100005],cnt=0;bool appear[100005];int main() {    while(1) {        memset(appear,0,sizeof appear);        memset(ru,0,sizeof ru);        bool fl=0,fst=1;        while(scanf("%d%d",&u,&v)&&u&&v) {            if(u<0&&v<0)                exit(0);            fst=0;            appear[u]=appear[v]=1;            ru[v]++;        }        if(fst) {            printf("Case %d is a tree.\n",++cnt);            continue;        }        int cnt0=0,ind;        for(int i=1;i<=100000;i++)            if(appear[i]&&ru[i]==0) {                cnt0++;                ind=i;                if(cnt0>1)                    break;            }        if(cnt0!=1)            fl=1;        for(int i=1;i<=100000;i++)            if(i==ind)                continue;            else if(ru[i]>1&&i!=ind) {                fl=1;                break;            }        printf("Case %d ",++cnt);        if(!fl)            puts("is a tree.");        else            puts("is not a tree.");    }    return 0;}

并查集做法

我自己没写..