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;
}