POJ 3280 Cheapest palindrome (interval dp)

Test instructions: Give a string, you want to put him into a palindrome, ask the minimum cost.

Solution: Initially thought only on both sides to add the letter, and then always WA. Later took someone else's program found can be in the middle of the letter.

DP, DP[I][J] represents the minimum cost of i-j string into palindrome strings.

Transfer equation: If the characters of I and J are the same, then the minimum cost of the strings they surround is directly. DP[I][J] = dp[i-1][j-1].

If it is not the same, then take the minimum value of the change on both sides. Dp[i][j] = min (Dp[i+1][j] + change[i], dp[i][j-1] + change[j]);

The code is as follows:

#include <iostream> #include <cstdio> #include <vector> #include <queue> #include <utility > #include <stack> #include <algorithm> #include <cstring> #include <string> #include <
cmath> #include <set> #include <map> using namespace std;
const int MAXN = 2e4 + 5;
int n, m;
Char STR[MAXN], ch[5];
int ADD[MAXN];

int DP[MAXN][MAXN]; int main () {#ifndef Online_judge freopen ("In.txt", "R", stdin),//Freopen ("OUT.txt", "w", stdout), #endif scanf ("%d
	%d ", &n, &m);
	scanf ("%s", str + 1);
		for (int i = 0, a, b; i < n; i++) {scanf ("%s%d%d", ch, &a, &b);
	Add[ch[0]] = min (a, b);
			} for (int k = 2, K <= m; k++) {for (int i = 1; I <= m; i++) {Int J = i + k-1; if (str[i] = = Str[j]) {Dp[i][j] = dp[i + 1][j-1];//If i+1 is greater than j-1 does not matter, the state that is not established is 0 cost} else {Dp[i][j] = min (dp[i + 1
			][J] + add[str[i]], dp[i][j-1] + add[str[j]);
	}}} printf ("%d\n", Dp[1][m]);
return 0; }