CF 477 C Dreamoon and Strings (DP)

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dp[i][j] 表示前i个字母里面去除j个字母后最多的不重复的p串的数量.
有两种情况，要么选取1~i中最后一个和p一样的串，要么不选取，dp[i][j] = max(dp[i-1][j], dp[ii][jj]+1), ii为匹配完p后在原串中的位置,jj为j-(匹配p过程中删去字母的数量);

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <vector>#include <cmath>#include <queue>#include <map>#include <set>using namespace std; #define INF 1000000000//typedef __int64 LL; #define N 2011#define M 511char str1[N], str2[N]; int dp[N][N]; int main() {#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin); #endif // ONLINE_JUDGEscanf("%s%s", str1+1, str2+1); int len1 = strlen(str1+1); int len2 = strlen(str2+1); for(int i = 1; i <= len1; i ++) {int ii = i, jj = len2; while(ii >= 1 && jj >= 1 ) {if(str1[ii] == str2[jj]) jj--; ii --; }if(jj == 0) {int dif = i - ii - len2; for(int th = 0; th <= ii; th ++) {dp[i][dif + th] = dp[ii][th] + 1; }}for(int th = 0; th < i; th ++) {dp[i][th] = max(dp[i][th], dp[i-1][th]); }}for(int i = 0; i <= len1; i ++) {printf("%d ", dp[len1][i]); }return 0; }

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