uva10739(String to Palindrome)

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思路：题目说的是给定一个字符串，然后可以对每个字符修改一次，问最少需要修改多少吃可以使得字符串变成回文字符串。这题的状态显然是很好像的dp[i][j]表示区间[i,j]的字符串修改成回文的需要修改的最少次数，然后就是转移方程了。

1，s[i] = s[j],dp[i][j] = dp[i + 1][j - 1];

2, s[i] != s[j],显然此事需要修改一次，看是修改s[i]还是修改s[j]，修改s[i]的话就是dp[i + 1][j],修改s[j]的话就是dp[i][j - 1],当然还有就是d[i + 1][j - 1];取其最小者 + 1;

/*****************************************Author      :Crazy_AC(JamesQi)Time        :2015File Name   :*****************************************/// #pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <algorithm>#include <iomanip>#include <sstream>#include <string>#include <stack>#include <queue>#include <deque>#include <vector>#include <map>#include <set>#include <stdio.h>#include <string.h>#include <math.h>#include <stdlib.h>#include <limits.h>using namespace std;#define MEM(a,b) memset(a,b,sizeof a)typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> ii;const int inf = 1 << 30;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;inline int Readint(){char c = getchar();while(!isdigit(c)) c = getchar();int x = 0;while(isdigit(c)){x = x * 10 + c - '0';c = getchar();}return x;}char s[1010];int dp[1010][1010];int main(){// freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);int t,icase = 0;scanf("%d",&t);while(t--){scanf("%s",s);int r = strlen(s);memset(dp, 0,sizeof dp);for (int i = r - 1;i >= 0;--i){for (int j = i + 1;j <= r - 1;++j){if (s[i] == s[j]) dp[i][j] = dp[i + 1][j - 1];else dp[i][j] = min(min(dp[i + 1][j],dp[i][j - 1]),dp[i + 1][j - 1]) + 1;}}printf("Case %d: %d\n",++icase,dp[0][r - 1]);}return 0;}

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