Codevs 2980 买帽子 dp(LIS)

来源：互联网发布：怎么取消备案域名编辑：程序博客网时间：2024/05/17 00:17

Codevs 2980 买帽子

对于一个字符串的最长对称子序列，就是它和它反串的最长公共子序列

我看了DQS的题解

代码：（char 数组实现）

#include <cstdio>#include <cstring>#include <iostream>#include <cstdlib>#include <algorithm>using namespace std;#define MAXLEN (500+5)#define MAXN (100+10)struct STR{    char s[MAXLEN];    int k, len;}c[MAXN];char zs[MAXLEN];int dp[MAXLEN][MAXLEN];bool cmp(STR a, STR b){    if(a.k != b.k) return a.k > b.k;    if(a.len != b.len) return a.len < b.len;    for(int i = 1; i <= a.len; i ++)        if(a.s[i] != b.s[i])            return a.s[i] < b.s[i];    return 0;}int main(){    int n;    cin >> n;    for(int q = 1; q <= n; q ++)    {        scanf("%s", c[q].s+1);        c[q].len = strlen(c[q].s+1);        for(int i = 1; i <= c[q].len; i ++)            zs[c[q].len-i+1] = c[q].s[i];        memset(dp, 0, sizeof(dp));        for(int i = 1; i <= c[q].len; i ++)            for(int j = 1; j <= c[q].len; j ++)                if(c[q].s[i] == zs[j])                    dp[i][j] = max(max(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1]+1);                else dp[i][j] = max(dp[i-1][j], dp[i][j-1]);        c[q].k = dp[c[q].len][c[q].len];    }    sort(c+1, c+n+1, cmp);    for(int i = 1; i <= n; i ++)    {        for(int j = 1; j <= c[i].len; j ++)            printf("%c", c[i].s[j]);        printf("\n");    }    return 0;}

0 0