UVA - 10617 Again Palindrome
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题意:只通过删除,问可以弄成多少的回文串,跟之前的一题是相似的,对于str[i] = str[j]的情况,如果我们步删除的话dp[i][j] = dp[i+1][j-1] +1,如果删除str[i]的话,dp[i][j] += dp[i+1][j],相似的还有dp[i][j] += dp[i][j-1],除去相同的dp[i+1][j-1],
#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MAXN = 70;char str[MAXN];long long dp[MAXN][MAXN];int main(){ int t; scanf("%d",&t); while (t--){ scanf("%s",str); memset(dp,0,sizeof(dp)); int len = strlen(str); for (int i = 0; i < len; i++) dp[i][i] = 1; for (int i = len - 2; i >= 0; i--) for (int j = i + 1; j < len; j++){ if (str[i] == str[j]) dp[i][j] = dp[i+1][j-1] + 1; dp[i][j] += dp[i+1][j] + dp[i][j-1] - dp[i+1][j-1]; } printf("%lld\n",dp[0][len-1]); } return 0;}
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