标签:int graph vertexes dfs Ai 寒假 maxn 搜索 Cycle
A tournament is a directed graph without self-loops in which every pair of vertexes is connected by exactly one directed edge. That is, for any two vertexes u and v (u ≠ v) exists either an edge going from u to v, or an edge from v to u.
You are given a tournament consisting of n vertexes. Your task is to find there a cycle of length three.
Input
The first line contains an integer n (1 ≤ n ≤ 5000). Next n lines contain the adjacency matrix A of the graph (without spaces). Ai, j = 1 if the graph has an edge going from vertex i to vertex j, otherwise Ai, j = 0. Ai, j stands for the j-th character in the i-th line.
It is guaranteed that the given graph is a tournament, that is, Ai, i = 0, Ai, j ≠ Aj, i (1 ≤ i, j ≤ n, i ≠ j).
Output
Print three distinct vertexes of the graph a1, a2, a3 (1 ≤ ai ≤ n), such that Aa1, a2 = Aa2, a3 = Aa3, a1 = 1, or "-1", if a cycle whose length equals three does not exist.
If there are several solutions, print any of them.
Examples
Input5Output
00100
10000
01001
11101
11000
1 3 2Input
5Output
01111
00000
01000
01100
01110
-1cycle
思路:
就是用搜索,以一排展开,dfs含有两个参数,这两个参数代表这个位置为1,所有,再去搜索,按照没有搜索的一排展开,
找到这个值。
AC之后,之前这个题目思路有一些混乱,再整理一下
在一个dfs要考虑到三个数,dfs里面的两个参数就是其中两个数,然后找到这两个参数中任意一个对应得第三个数,
再判断第三个数和另一个参数有没有相互对应,有就返回了,没有继续判断这一行有没有被标记,标记了就不用,没有标记就
继续进行搜索。值得确定得是,第一确实每一个都搜到了,没有遗漏,第二,就是搜索得递归,可以再仔细想想
#include <stdio.h> #include <stdlib.h> #include <string.h> using namespace std; const int maxn=5050; char ana[maxn][maxn]; bool bna[maxn][maxn],vis[maxn]; int n,a,b,c; int dfs(int x,int y) { vis[x]=1; for(int i=1;i<=n;i++) { if(bna[x][i]) { if(y&&bna[i][y]) { a=y; b=x; c=i; return 1; } if(!vis[i]) if(dfs(i,x)) return 1; } } return 0; } int main() { scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%s",ana[i]+1); for(int j=1;j<=n;j++) { bna[i][j]=ana[i][j]-'0'; } } for(int i=1;i<=n;i++) { if(!vis[i]) { if(dfs(i,0)) { printf("%d %d %d\n",a,b,c); return 0; } } } printf("-1\n"); return 0; }
标签:int,graph,vertexes,dfs,Ai,寒假,maxn,搜索,Cycle 来源: https://www.cnblogs.com/EchoZQN/p/10351489.html
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