UVa 1630 Folding (区间DP)

题意：折叠一个字符串，使得其成为一个尽量短的字符串  例如AAAAAA变成6（A）

而且这个折叠是可以嵌套的，例如 NEEEEERYESYESYESNEEEEERYESYESYES 会变成 2(N5(E)R3(YES))。

析：用dp[i][j] 表示字符串中的第 i 个到第 j 个字符压缩后的最短长度。那么就有两种方式，一种就是自身压缩都最短，另一种就是两段分别压缩，

然后再接起来最短。

代码如下：

 

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e2 + 5;
const LL mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
string str;
string dp[maxn][maxn];

int cal(int l, int r){
    int len = (r - l + 1);
    for(int i = 1; i <= len/2; ++i) if(len % i == 0){
        bool ok = true;
        for(int j = l; j <= r-i; ++j)  if(str[j] != str[j+i]){
            ok = false;  break;
        }
        if(ok)  return i;
    }
    return 0;
}

void solve(){
    for(int i = 0; i < n; ++i)  dp[i][i] = str[i];
    for(int i = n-2; i >= 0; --i){
        for(int j = i+1; j < n; ++j){
            int ans = INF, x;
            for(int k = i; k < j; ++k) if(ans > dp[i][k].size() + dp[k+1][j].size()){
                ans = dp[i][k].size() + dp[k+1][j].size();
                x = k;
            }
            dp[i][j] = dp[i][x] + dp[x+1][j];
            int len = cal(i, j);
            if(len){
                char s[5];
                sprintf(s, "%d", (j-i+1)/len);
                string tmp = (string)s + "(" + dp[i][i+len-1] + ")";
                if(tmp.size() <= ans)  dp[i][j] = tmp;
            }
        }
    }
}

int main(){
    while(cin >> str){
        n = str.size();
        solve();
        cout << dp[0][n-1] << endl;
    }
    return 0;
}