經典區間DP。
首先添加一個字符和刪除一個字符是等價的,因爲在一個位置添加一個字符,就等價與在對稱迴文的位置刪除一個字符,刪除同理。那麼我們只需要考慮刪除字符。
設dp[l][r]表示將[l, r]改爲迴文串的最小代價,那麼有
(1)dp[l][r] = min(dp[l + 1][r] + cost[str[l]], dp[l][r - 1] + cost[str[r]])
(2)如果str[l] == str[r],那麼dp[l][r] = min(dp[l][r], dp[l + 1][r - 1])
/* Telekinetic Forest Guard */
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 2005, maxm = 28, inf = 0x3f3f3f3f;
int n, m, num[maxn], dp[maxn][maxn], cost[maxm];
char str[maxn];
inline int iread() {
int f = 1, x = 0; char ch = getchar();
for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
return f * x;
}
inline int dfs(int l, int r) {
if(l == r) return 0;
if(~dp[l][r]) return dp[l][r];
int res = inf;
res = min(dfs(l + 1, r) + cost[num[l]], dfs(l, r - 1) + cost[num[r]]);
if(num[l] == num[r]) res = min(res, dfs(l + 1, r - 1));
return dp[l][r] = res;
}
int main() {
m = iread(); n = iread();
scanf("%s", str + 1);
for(int i = 1; i <= n; i++) num[i] = str[i] - 'a' + 1;
for(int i = 1; i <= m; i++) {
int w1, w2;
scanf("%s%d%d", str, &w1, &w2);
cost[str[0] - 'a' + 1] = min(w1, w2);
}
for(int i = 1; i <= n; i++) for(int j = i; j <= n; j++) dp[i][j] = -1;
printf("%d\n", dfs(1, n));
return 0;
}