poj 2176 folding

Description：

  就是把一个字符串压尽可能的压缩

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 using namespace std;
 const int N = , INF = 1e9;
 struct node{
     char s[N];  //dp[i][j]的len表示区间i-j压缩后的最小长度 s表示压缩后的字串
     int len;
 }dp[N][N];
 int n;
 char s[N];
 int main(){
     scanf("%s", s + );
     n = strlen(s + );
     for(int i = ; i <= n; i++)
       dp[i][i].len = , dp[i][i].s[] = s[i];
     for(int l = ; l <= n; l++)
       for(int i = ; i + l -  <= n; i++){ //枚举区间
           int j = i + l - ;
           dp[i][j].len = INF;
           for(int nowl = ; nowl <= l / ; nowl++){  //枚举一个区间所有可能被压缩后子串的长度
               if(l % nowl != ) continue;  //如果无法整除肯定不能被压缩
               int st = i, ed = i + nowl;
               while(s[st] == s[ed] && ed <= j) st++, ed++;  //检验该串能否被压缩
               if(ed > j){
                   int num = l / nowl;  //压缩后串的数目为num
                   sprintf(dp[i][j].s, "%d", num);  //那么该状态下最小串的开头就是num
                   strcat(dp[i][j].s, "(");
                   strcat(dp[i][j].s, dp[i][i + nowl - ].s); //把压缩后的串接起来
                   strcat(dp[i][j].s, ")");
                   dp[i][j].len = strlen(dp[i][j].s);
                   break;
               }
           }
           for(int k = i; k < j; k++)
             if(dp[i][j].len > dp[i][k].len + dp[k + ][j].len){  //重新扫一遍区间更新dp[i][j]
                 dp[i][j].len = dp[i][k].len + dp[k + ][j].len;
                 strcpy(dp[i][j].s, dp[i][k].s);
                 strcat(dp[i][j].s, dp[k + ][j].s);
             }
       }
     puts(dp[][n].s);
     return ;
 }