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UVa 10739 String to Palindrome (DP)2014-07-29 csdn博客 synapse7

10739 - String to Palindrome

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=114&page=show_problem&problem=1680

思路:对于每个区间[i, j]:

若str[i] == str[j],dp[i][j] = dp[i + 1][j - 1];

若str[i] != str[j], dp[i][j] = min(dp[i + 1][j], dp[i] [j - 1], dp[i + 1] [j - 1]) + 1 (删去i,删去j,修改i为j(添加与删除是一样的))

完整代码:

/*0.045s*/#include<bits/stdc++.h>using namespace std;const int N = 1005;int dp[N][N];char str[N];int main(){int cas, t = 0, len,i;scanf("%d
", &cas);while (cas--){gets(str);len = strlen(str);memset(dp, 0, sizeof(dp));for (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];///向内考虑else dp[i][j] = min(min(dp[i + 1][j], dp[i][j - 1]), dp[i + 1][j - 1]) + 1;///删去i,删去j,修改i为j(添加与删除是一样的)}}printf("Case %d: %d
", ++t, dp[0][len - 1]);}return 0;}