poj 2138 DFS

来源：互联网发布：神结局电影知乎编辑：程序博客网时间：2024/06/05 16:34

首先预处理，哪些字符串是可以互相到达的

还有一点要注意的是，如果一个都找不出来，就输出原单词

AC代码如下：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;struct Edge{    int to, next;};char s[1100][100];int N;Edge edge[400000];int head[1100], tot;int dp[1100];int anspos, ans;void add_edge( int a, int b ){    edge[tot].to = b;    edge[tot].next = head[a];    head[a] = tot++;}bool judge( char a[], char b[] ){    if( strlen( a ) != strlen( b ) - 1 ){        return false;    }    int i, j;    i = j = 0;    while( a[i] && b[j] ){        if( a[i] == b[j] ){            i++;            j++;        }else{            j++;        }        if( j - i > 1 ){            break;        }    }    if( j - i > 1 ){        return false;    }    if( (unsigned)i == strlen( a ) && (unsigned)j == strlen( b ) ){        return true;    }    if( (unsigned)i == strlen( a ) && (unsigned)j == strlen( b ) - 1 ){        return true;    }    return false;}int DFS( int pos, int deep ){    if( deep > ans ){        anspos = pos;        ans = deep;    }    for( int i = head[pos]; i != -1; i = edge[i].next ){        DFS( edge[i].to, deep + 1 );    }    return 0;}int main(){    while( scanf( "%d%s", &N, s[0] ) != EOF ){        tot = 0;        memset( head, -1, sizeof( head ) );        for( int i = 1; i <= N; i++ ){            scanf( "%s", s[i] );        }        for( int i = 0; i <= N; i++ ){            for( int j = 1; j <= N; j++ ){                if( i != j ){                    if( judge( s[i], s[j] ) ){                        add_edge( i, j );                    }                }            }        }        anspos = 0;        ans = 0;        DFS( 0, 0 );        if( ans == 0 ){            cout << s[0] << endl;        }else{            cout << s[anspos] << endl;        }    }    return 0;}

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