codeforces B. Strongly Connected City(dfs水过)
2014-10-06 09:10
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题意:有横向和纵向的街道,每个街道只有一个方向,垂直的街道相交会产生一个节点,这样每个节点都有两个方向,
问是否每一个节点都可以由其他的节点到达....
思路:规律没有想到,直接爆搜!每一个节点dfs一次,记录每个节节点被访问的次数!如果每个节点最终的访问次数
和所有节点的数目相同,则输出“YES", 否则输出”NO“
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问是否每一个节点都可以由其他的节点到达....
思路:规律没有想到,直接爆搜!每一个节点dfs一次,记录每个节节点被访问的次数!如果每个节点最终的访问次数
和所有节点的数目相同,则输出“YES", 否则输出”NO“
#include <queue> #include <string> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; char s1[25], s2[25]; int vis[25][25]; int cnt[25][25]; int n, m; void dfs(int x, int y){ if(x<1 || x>n || y<1 || y>m || vis[x][y]) return; ++cnt[x][y]; vis[x][y] = 1; if(s1[x] == '<') dfs(x, y-1); else dfs(x, y+1); if(s2[y] == 'v') dfs(x+1, y); else dfs(x-1, y); } int main(void) { scanf("%d%d", &n, &m); scanf("%s%s", s1+1, s2+1); for(int i=1; i<=n; ++i) for(int j=1; j<=m; ++j){ memset(vis, 0, sizeof(vis)); dfs(i, j); } int s = n*m; for(int i =1; i<=n; ++i) for(int j=1; j<=m; ++j) if(cnt[i][j] != s){ cout<<"NO"<<endl; return 0; } cout<<"YES"<<endl; return 0; }
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