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题目链接:http://poj.org/problem?id=3280
这道题首先要清楚:对于构成一个回文串,删去一个字符和加上一个字符是等效的,所以我们取花费较少的情况。
转移方程为:dp[i][j] = dp[i-1][j-1](s[i]==s[j])因为已经构成回文串,并且dp[i-1][j-1]是最优的。
dp[i][j] = min(dp[i][j], dp[i + 1][j] + use[s[i] - 'a']) ——左边
dp[i][j] = min(dp[i][j], dp[i][j - 1] + use[s[j] - 'a']) ——右边
AC代码:
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<algorithm> 5 6 using namespace std; 7 8 char s[2005]; 9 int use[2005], dp[2005][2005]; 10 11 int main(){ 12 int n, m, x, y; 13 char c; 14 scanf("%d%d", &n, &m); 15 scanf("%s", s + 1); 16 for(int i = 1; i <= n; i++){ 17 cin >> c >> x >> y; 18 if(x < y) use[c - 'a'] = x; 19 else use[c - 'a'] = y; 20 } 21 for(int l = 1; l <= m; l++){ 22 for(int i = 1; i <= m; i++){ 23 int j = i + l; 24 if(j > m) break; 25 dp[i][j] = 0x3f3f3f; 26 if(s[i] == s[j]) dp[i][j] = dp[i + 1][j - 1]; 27 else{ 28 dp[i][j] = min(dp[i][j], dp[i + 1][j] + use[s[i] - 'a']); 29 dp[i][j] = min(dp[i][j], dp[i][j - 1] + use[s[j] - 'a']); 30 } 31 } 32 } 33 printf("%d\n", dp[1][m]); 34 return 0; 35 }AC代码