uva1630
題目描述：給出一個由大寫字母組成的長度為n（1≤n≤100）的串，“折疊”成一個盡量短的串。例 如，AAAAAAAAAABABABCCD折疊成9(A)3(AB)CCD。折疊是可以嵌套的，例 如，NEERCYESYESYESNEERCYESYESYES可以折疊成2(NEERC3(YES))。多解時可以輸出 任意解。
分析：枚舉分割字符串位置，檢查是否有字串再和原串長度比較。
#include<iostream> #include<algorithm> #include<math.h> #include<string.h> #include<stdio.h> #include<string> #include<vector> #include<queue> #include<map> #include<sstream> #include<cassert> #include<set> //#include <bits/stdc++.h> using namespace std; typedef long long ll; const int INF = 0x3f3f3f3f; const int maxn = 100+10; int dp[maxn][maxn]; string sub[maxn][maxn]; string str; int check(int l, int r) {for (int i = 1; i <= (r - l + 1) / 2; i++) {int ok = 1;if ((r - l + 1) % i)continue;for (int j = l; j + i <= r; j++) {if (str[j] != str[j + i]) {ok = 0;break;}}if (ok)return i;}return 0; }int solve(int l, int r) {int &ans = dp[l][r];if (ans != -1)return ans;if (l == r) {sub[l][r] = str[l];return ans = 1;}ans = 1;int res = INF, k;for (int i = l; i < r; i++) {int t = solve(l, i) + solve(i + 1, r);if (t < res) {k = i;res = t;}}sub[l][r] = sub[l][k] + sub[k + 1][r];int len = check(l, r);if (len) {int o = (r - l + 1) / len;stringstream ss;ss << o;string t = ss.str();string astr = t + "(" + sub[l][l + len - 1] + ")";int t1 = astr.size();if (t1 < res) {res = t1;sub[l][r] = astr;}}return ans = res; }int main() {while (cin>>str) {memset(dp, -1, sizeof(dp));int n = str.size() - 1;solve(0, n);cout << sub[0][n] << endl;}return 0; }?
總結
