• Codeforces 323 B Tournament-graph


    Discription

    In this problem you have to build tournament graph, consisting of n vertices, such, that for any oriented pair of vertices (v, u(v ≠ u) there exists a path from vertexv to vertex u consisting of no more then two edges.

    A directed graph without self-loops is a tournament, if there is exactly one edge between any two distinct vertices (in one out of two possible directions).

    Input

    The first line contains an integer n (3 ≤ n ≤ 1000), the number of the graph's vertices.

    Output

    Print -1 if there is no graph, satisfying the described conditions.

    Otherwise, print n lines with n integers in each. The numbers should be separated with spaces. That is adjacency matrix a of the found tournament. Consider the graph vertices to be numbered with integers from 1 to n. Then av, u = 0, if there is no edge from v to u, and av, u = 1 if there is one.

    As the output graph has to be a tournament, following equalities must be satisfied:

    • av, u + au, v = 1 for each v, u (1 ≤ v, u ≤ nv ≠ u);
    • av, v = 0 for each v (1 ≤ v ≤ n).

    Example

    Input
    3
    Output
    0 1 0
    0 0 1
    1 0 0
    Input
    4
    Output
    -1


    构造题,比较迷。
    首先我们如果有了n个点的合法图是很容易构造出n+2个点的合法图的。
    首先可以把1的入点看成一类点,1的出点看成一类点,然后再加上1本身和n+1和n+2,我们现在就有了5个点。
    而我们的目的是让任意一对点都在至少一个三元环 (形如a->b,b->c,c->a) 出现。
    所以直接xjb构造就行了,这个其实可以不用手算,直接写个程序跑一跑也是可以的2333(我是不会告诉你们我的连边方案就是电脑枚举出来的哈哈)

    但是发现4没有答案,导致我一开始以为偶数都是gg的然后就WA了。。。
    后来写了个搜索发现6是有答案的2333,所以是偶数的话特判完了之后直接从6的图往后跑就行了,n=6的图可以搜索也可以手玩(反正写搜索就当练暴力了2333)
    #include<bits/stdc++.h>
    #define ll long long
    #define maxn 1005
    using namespace std;
    int a[maxn][maxn];
    int n,m,k,tone[maxn];
    int main(){
    	scanf("%d",&n);
    	if(!(n&1)){
    		if(n<6){
    			puts("-1");
    			return 0;
    		}
    		
    		a[1][5]=a[1][6]=1;
    		a[2][1]=a[2][6]=1;
    		a[3][1]=a[3][2]=a[3][5]=1;
    		a[4][1]=a[4][2]=a[4][3]=1;
    		a[5][2]=a[5][4]=1;
    		a[6][3]=a[6][4]=a[6][5]=1;
    		
    		tone[2]=tone[3]=tone[4]=1;
    		
    		for(int i=7;i<=n;i+=2){
    			tone[i]=1;
    			a[i][1]=1,a[1][i+1]=1,a[i+1][i]=1;
    			for(int j=2;j<i;j++){
    				if(tone[j]) a[j][i]=1,a[i+1][j]=1;
    				else a[j][i+1]=1,a[i][j]=1;
    			}
    		}
    	}
    	else{
    		a[1][2]=a[2][3]=a[3][1]=1;
    		tone[3]=1;
    		for(int i=4;i<=n;i+=2){
    			tone[i]=1;
    			a[i][1]=1,a[1][i+1]=1,a[i+1][i]=1;
    			for(int j=2;j<i;j++){
    				if(tone[j]) a[j][i]=1,a[i+1][j]=1;
    				else a[j][i+1]=1,a[i][j]=1;
    			}
    		}
    	}	
    
    	for(int i=1;i<=n;i++){
    		for(int j=1;j<=n;j++) printf("%d ",a[i][j]);
    		puts("");
    	}
    
    	
    	/*
    	for(int i=1;i<=n;i++)
    	    for(int j=1;j<=n;j++) a[i][j]=(a[i][j]?1:1<<30);
    	for(int i=1;i<=n;i++) a[i][i]=0;
    	
    	for(int k=1;k<=n;k++)
    	    for(int i=1;i<=n;i++)
    	        for(int j=1;j<=n;j++) if(a[i][k]+a[k][j]<a[i][j]) a[i][j]=a[i][k]+a[k][j];
    	        
    	for(int i=1;i<=n;i++)
    	    for(int j=1;j<=n;j++) if(a[i][j]>2) puts("No");
    	*/
    	
    	return 0;
    }
    
    
    

      

     
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  • 原文地址:https://www.cnblogs.com/JYYHH/p/8535153.html
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