题意:给你一个由小写字母组成的字符串s,你可以在两端添加或删除字母使其变为回文串,求最小花费。
思路:区间dp。用dp[l][r]表示l到r变为回文串的最小花费。则dp[l][r] = min(dp[l+1][r] + add[l], dp[l + 1][r] + del[l],dp[l][r - 1] + add[r], dp[l][r - 1] + del[r]).如果s[l] == s[r], 那么dp[l][r] = dp[l+1][r-1].
#include <cstdio>
#include <algorithm>
#include <iostream>
#include<vector>
#include<cmath>
#include<set>
#include<cstring>
#include<map>
using namespace std;
typedef long long ll;
const int maxn = 21e2 + 10;
const int maxt = 100200;
const int inf = 0x3f3f3f3f;
const ll INF = 0x7f7f7f7f7f;
const int mod = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
int n, m;
char s[2010];
int a[27], b[27];
int dp[2010][2010];
int DP(int l, int r){
int &ans = dp[l][r];
if(l >= r) return ans = 0;
if(ans != -1) return ans;
if(s[l] == s[r]) return ans = DP(l + 1, r - 1);
ans = inf;
int t1 = s[l] - 'a', t2 = s[r] - 'a';
ans = min(ans, DP(l, r - 1) + b[t2]);
ans = min(ans, DP(l + 1, r) + b[t1]);
ans = min(ans, DP(l, r - 1) + a[t2]);
ans = min(DP(l + 1, r) + a[t1], ans);
return ans;
}
int main(){
scanf("%d%d", &n, &m);
scanf("%s", s);
for(int i = 0; i < n; ++i){
char ss[2];
int x, y;
scanf("%s%d%d", ss, &x, &y);
a[ss[0] - 'a'] = x;
b[ss[0] - 'a'] = y;
}
memset(dp, -1, sizeof dp);
printf("%d\n", DP(0, m - 1));
return 0;
}