HDU 1560

题意

给出n( 1 <= n <= 8 )个长度为1~5的DNA序列, 找出一个最短的满足包含所有这些子字符串的字符串，只要字符串中各字母相对位置相同即可。

思路

IDA*
从maxd = lenmax开始加深

AC代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define mst(a) memset(a, 0, sizeof(a));

using namespace std;
int maxd, n;
char g[10][6];
int len[10], pos[10];
char DNA[] = "ACGT";
bool flag;

void init(){
    flag = false;
    mst(g);
    mst(len);
    mst(pos);
}

void dfs( int d ){
    int result = 0;
    for(int i = 0; i < n; i++)
        result = max(result, len[i]-pos[i]);
    if( result == 0 ){ flag = true;  return; }
    else if( result > d )  return ;  //剪枝
    int _pos[10]; //用于记录改变前状态
    for( int i = 0; i < n; i++ )
        _pos[i] = pos[i];
    bool ok = false;
    for( int i = 0; i < 4; i++ ){
        for( int j = 0; j < n; j++ )
            if( g[j][pos[j]] == DNA[i] ){
                pos[j]++;
                ok = true;
            }
        if(ok){
            dfs(d-1);
            if( flag )  return;
            for( int k = 0; k < n; k++ ) //若未成功则还原pos串
                pos[k] = _pos[k];  
        }
    }
}

void solve(){
    for( ; ; maxd++ ){
        dfs(maxd);
        if( flag )
            break;
    }
    printf("%d
",maxd);
}

int main()
{
    int T;
    scanf("%d",&T);
    while( T-- ){
        init();
        int lenmax = -1;
        scanf("%d", &n);
        for( int i = 0; i < n; i++ ){
            scanf("%s",g[i]);
            len[i] = strlen(g[i]);
            pos[i] = 0;
            lenmax = max(lenmax, len[i]);
        }
        maxd = lenmax;  //从lenmax开始枚举
        solve();
    }
    return 0;
}