MUTC 3 E - Triangle LOVE 图论/搜索

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1629    Accepted Submission(s): 693

Problem Description

Recently, scientists find that there is love between any of two people. For example, between A and B, if A don’t love B, then B must love A, vice versa. And there is no possibility that two people love each other, what a crazy world!
Now, scientists want to know whether or not there is a “Triangle Love” among N people. “Triangle Love” means that among any three people (A,B and C) , A loves B, B loves C and C loves A.
  Your problem is writing a program to read the relationship among N people firstly, and return whether or not there is a “Triangle Love”.

 

Input

The first line contains a single integer t (1 <= t <= 15), the number of test cases.
For each case, the first line contains one integer N (0 < N <= 2000).
In the next N lines contain the adjacency matrix A of the relationship (without spaces). A_i,j = 1 means i-th people loves j-th people, otherwise A_i,j = 0.
It is guaranteed that the given relationship is a tournament, that is, A_i,i= 0, A_i,j ≠ A_j,i(1<=i, j<=n,i≠j).

 

Output

For each case, output the case number as shown and then print “Yes”, if there is a “Triangle Love” among these N people, otherwise print “No”.
Take the sample output for more details.

 

Sample Input

2
5
00100
10000
01001
11101
11000
5
01111
00000
01000
01100
01110

 

Sample Output

Case #1: Yes
Case #2: No

 

Author

BJTU

 

Source

2012 Multi-University Training Contest 3

 

Recommend

zhoujiaqi2010

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图论做法:

一个结论:竞赛图上只要有环,就有三元环。

搜索做法:

带上父节点直接搜= =

标准做法1:

	增量算法，充分利用“任意两点间仅有一条有向边”的性质。
	假设前面已经加入了N个点，现在来了第N+1个点。
	那么一定能将N个点分成left和right两部分，使得N+1号点到left有边，right到N+1号点右边（因为任意两点间都有边），那么，如果left的任意一个点l到right任意一个点r有边的话，那么就有答案N+1->l->r->N+1这样一个长度为3的环。
	那么每次加入N+1号点后，用O(N)的复杂度求出左边的数量leftnum，右边的数量rightnum，left的出度和leftout，left的入度和leftin。
	如果left没有一条到right的边，则一定满足：
	leftin = leftout + leftnum * rightnum（left和right任意两点右边，如果没有左到右的，那么leftnum*rightnum条边都是右到左的）
	那么，如果leftin != leftout + leftnum * rightnum，则暴力枚举左点，右点即可得到答案。
	总体复杂度O(n^2)

标准做法2:

	归结为1个条件，任何两人都有边，要么出要么入。
	对于有向三元环，我们知道找到：A->B->C->A ，B->C->A->B ，C->A->B->C 是一样的，这给0(n^2)的算法提供了基础。
        对于每次枚举i，我们在(0~i-1)的范围看下有多少个i指向的点(剩下的就是指向i的点)，同时算下i指向的点的出度和。就可以知道这些 i指向的点 指向 指向i的点（剩下的点）的数目，如果 
        num * (num - 1) / 2 < sumout 那么就是被指向的点有指出去了，这样就形成了3元环。
        否则，剩下的就是更新下出度即可，继续执行下一个节点。
 

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int maxn=2222;

char s[maxn][maxn];

int inq[maxn];
int n;

bool top_sort()
{
    int num=n;
    queue<int>que;
    for (int i=0;i<n;i++)
        if (!inq[i]) {que.push(i);num--;}
    while (!que.empty())
    {
        int top=que.front();
        que.pop();
        for (int i=0;i<n;i++)
        {
            if (top!=i&&s[top][i]=='1')
            {
                inq[i]--;
                if (!inq[i]) {que.push(i);num--;}
            }
        }
    }
    if (num) return true;
    else return false;

}

int main()
{
    int T,cnt=0;;
    scanf("%d",&T);
    while (T--)
    {
        memset(inq,0,sizeof(inq));
        scanf("%d",&n);
        for (int i=0;i<n;i++)
        {
            scanf("%s",s[i]);
        }
        for (int i=0;i<n;i++)
            for (int j=0;j<n;j++)
                if (i!=j&&s[i][j]=='1') inq[j]++;
        bool ret=top_sort();
        printf("Case #%d: ",++cnt);
        if (ret) puts("Yes");
        else puts("No");
    }
    return 0;
}

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#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;
bool v[5555]= {0};
int n;
char s[5555][5555];
bool dfs(int i,int dad)
{
    v[i]=true;
    for (int j=1; j<=n; j++)
        if (s[i][j]-'0')
        {
            if (s[j][dad]-'0') return true;
            if (!v[j]) if (dfs(j,i)) return true;
        }
    return false;
}

int main()
{
    int T;
    scanf("%d",&T);
    for (int lp=1; lp<=T; lp++)
    {
        memset(v,0,sizeof(v));
        scanf("%d",&n);
        for (int i=1; i<=n; i++)
        {
            scanf("%s",s[i]+1);
        }
        printf("Case #%d: ",lp);
        bool flag=false;
        for (int i=1; i<=n; i++)
        {
            if (!v[i])
            {
                if (dfs(i,i)){flag=true;break;}
            }
        }
        if (!flag) printf("No\n");
        else printf("Yes\n");
    }
    return 0;
}