题目连接：http://lightoj.com/volume_showproblem.php?problem=1033
题目大意：给你一个字符串，问至少添加几个字符串，才能使这个字符串变成回文串
解题思路：用dp[i][j]表示[i,j]内的字符需要添加几个字符才能变回文
考虑两种情况
1.str[i] == str[j]，那么dp[i][j] = dp[i +1][j - 1]
2.str[i] != str[j]，那么只能在左边添加一个，或者右边添加一个了
所以dp[i][j] = min(dp[i + 1][j], dp[i][j - 1]) + 1
1 #include <algorithm> 2 #include <iostream> 3 #include <cstring> 4 #include <cstdlib> 5 #include <cstdio> 6 #include <vector> 7 #include <ctime> 8 #include <queue> 9 #include <list> 10 #include <set> 11 #include <map> 12 using namespace std; 13 #define INF 0x3f3f3f3f 14 typedef long long LL; 15 16 char a[105]; 17 int dp[110][110]; 18 int dfs(int l, int r) 19 { 20 if(l >= r) 21 return 0; 22 if(dp[l][r] >= 0) 23 return dp[l][r]; 24 int sum = INF; 25 for(int k = l+1; k <= r; k++) 26 { 27 if(a[l] == a[k]) 28 sum = min(sum, dfs(l+1, k-1) + r - k); 29 else 30 sum = min(sum, min(dfs(l+1, k), dfs(l, k-1))+1+r-k); 31 } 32 dp[l][r] = sum; 33 return sum; 34 } 35 int main() 36 { 37 int t; 38 scanf("%d", &t); 39 for(int ca = 1; ca <= t; ca++) 40 { 41 scanf("%s", a); 42 int len = strlen(a); 43 memset(dp, -1, sizeof(dp)); 44 int sum = dfs(0, len-1); 45 46 printf("Case %d: %d ", ca, sum); 47 } 48 return 0; 49 }