Compress String

時(shí)間限制：2000?ms ?|? 內(nèi)存限制：65535?KB 難度：3 描述

One day,a beautiful girl ask LYH to help her complete a complicated task—using a new compression method similar to Run Length Encoding(RLE) compress a string.Though the task is difficult, LYH is glad to help her. The compress rules of the new method is as follows: if a substring S is repeated k times, replace k copies of S by k(S). For example,?letsgogogo?is compressed into?lets3(go). The length of?letsgogogo?is 10, and the length of?lets3(go)?is 9. In general, the length of k(S) is (number of digits in k ) + (length of S) + 2. For example, the length of?123(abc)?is 8. It is also possible that substring?S is a compressed string. For example?nowletsgogogoletsgogogoandrunrunrun?could be compressed as?now2(lets3(go))and3(run). In order to make the girl happy, LYH solved the task in a short time. Can you solve it? 輸入

Thera are multiple test cases.
Each test case contains a string, the length of the string is no more than 200, all the character is lower case alphabet.

輸出

For each test case, print the length of the shortest compressed string.

樣例輸入

ababcd letsgogogo nowletsgogogoletsgogogoandrunrunrun

樣例輸出

6 9 24

題意：給出一個(gè)長(zhǎng)度不超過(guò)200的字符串，把這個(gè)字符串按照一定規(guī)則壓縮，即可以把幾個(gè)連續(xù)的相同子串壓縮成一個(gè)串，例如可以把letsgogogo壓縮為lets3(go)，壓縮后的子串如果還可以繼續(xù)壓縮，則可以繼續(xù)壓縮，如可以將nowletsgogogoletsgogogoandrunrunrun壓縮為now2(lets3(go))and3(run)。問(wèn)經(jīng)過(guò)壓縮后這個(gè)字符串的最短長(zhǎng)度是多少。

分析：?區(qū)間DP,dp[i][j]表示從i到j(luò)的字符串表示的最短長(zhǎng)度。

? ? ? ? ? dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j])。
? ? ? ? ? 然后去判斷當(dāng)前子串能不能壓縮，即是否由重復(fù)字符串組成，判斷時(shí)只需暴力枚舉重復(fù)長(zhǎng)度，去判斷即可。
? ? ? ? ? 如果當(dāng)前子串可以壓縮,則dp[i][j] = min(dp[i][j], dp[i][i + len - 1] + 2 + digcount((j - i + 1) / len));，

? ? ? ? ? 注意如果是數(shù)字，要用數(shù)字的位數(shù)表示增加的個(gè)數(shù)，而不是單純的增加1.

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 210; #define INF 0x3fffffff char str[N]; int n, dp[N][N];int digit_cnt(int x) {int a = 0;while(x) {a++;x /= 10;}return a; }bool check(int l, int r, int len) {if((r - l + 1) % len) return false;for(int i = l; i < l + len; i++) {for(int j = i + len; j <= r; j += len)if(str[i] != str[j]) return false;}return true; }int get_ans() {int i, j, k;n = strlen(str+1);for(i = 1; i <= n; i++) dp[i][i] = 1;for(i = n - 1; i >= 1; i--) {for(j = i + 1; j <= n; j++) {dp[i][j] = INF;for(k = i; k < j; k++)dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j]);for(int len = 1; len <= j-i+1; len++) {if(check(i, j, len)) {dp[i][j] = min(dp[i][j], dp[i][i+len-1] + 2 + digit_cnt((j - i + 1) / len));}}}}return dp[1][n]; }int main() {while(~scanf("%s", str+1)) {printf("%d\n", get_ans());}return 0; }

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