UVA - 1630 Folding(串折叠)(dp---记忆化搜索)

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 100 + 10;
const int MAXT = 10000 + 10;
using namespace std;
string s;
string ans[MAXN][MAXN];
int dp[MAXN][MAXN];
int dfs(int l, int r){
    if(dp[l][r] != -1) return dp[l][r];
    int len = r - l + 1;
    if(len == 1){//串的长度为1，不能折叠也不能枚举分割线
        ans[l][r] = s[l];
        return dp[l][r] = 1;
    }
    ans[l][r] = s.substr(l, len);
    int tmp = len;//以下判断串是否能折叠
    for(int i = 1; i <= len / 2; ++i){//枚举循环周期的长度
        if(len % i) continue;
        bool ok = true;
        for(int j = l + i; j <= r; j += i){//判断串是否以周期为i循环
            for(int k = 0; k < i; ++k){
                if(s[l + k] != s[j + k]){
                    ok = false;
                    break;
                }
            }
            if(!ok) break;
        }
        if(ok){//该串可以按周期为i折叠
            char t[10];
            sprintf(t, "%d", len / i);//循环串的长度
            dfs(l, l + i - 1);//循环串自身可能是可折叠的
            string str(t);
            str += "(" + ans[l][l + i - 1] + ")";
            int nowlen = (int)str.size();
            if(nowlen < tmp){//若折叠后的长度小于不折叠，则更新ans[l][r]
                tmp = nowlen;
                ans[l][r] = str;
            }
        }
    }
    if(tmp != len) return dp[l][r] = tmp;//如果可折叠
    for(int i = l; i < r; ++i){//该串不可折叠，枚举分割线
        int x = dfs(l, i);
        int y = dfs(i + 1, r);
        if(x + y < tmp){
            tmp = x + y;
            ans[l][r] = ans[l][i] + ans[i + 1][r];
        }
    }
    return dp[l][r] = tmp;
}
int main(){
    while(cin >> s){
        memset(dp, -1, sizeof dp);
        int len = (int)s.size();
        dfs(0, len - 1);
        printf("%s\n", ans[0][len - 1].c_str());
    }
    return 0;
}