POJ 3280 Cheapest Palindrome（有意思的區間dp）

題目鏈接

題意：
給你一個字符串，現在將某個字符增加或者減少都分別有一個代價，現在問你要讓這個人字符串變爲迴文串，最小的代價是什麼？

這題的關鍵是要發現刪除一個字符和增加一個字符是一樣的。比如說你增加了一個字符，使他變爲迴文串，那這個字符在另一端肯定也有一個與之相同的字符，那麼把他們刪了，還是迴文串，所以對一個字符進行操作的最小代價就是min(增加的代價，減少的代價)，所以增加某個字符c和刪除某個字符c的效果是一樣的。
dp[i][j]:使字符串區間[i,j]這部分變爲迴文串的最小花費
然後區間dp
1.先預處理一下dp[i][i] = 0 (1 <= i <= m)
2.如果s[i] == s[j]，dp[i][j] = dp[i - 1][j - 1];
3.否則，dp[i][j] = min(dp[i + 1][j] + cost[i], dp[i][j - 1] + cost[j]);

#include  <map>
#include  <set>
#include  <cmath>
#include  <queue>
#include  <cstdio>
#include  <vector>
#include  <climits>
#include  <cstring>
#include  <cstdlib>
#include  <iostream>
#include  <algorithm>
#include  <time.h>
#include  <sstream>

#define eb emplace_back
#define mt make_tuple
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, a, b) for (int i = (int)(a); i >= (int)b; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)
#define rep(i, l, r) for (int i = (l); i <= (r); i++)
#define per(i, r, l) for (int i = (r); i >= (l); i--)
#define ms(x, y) memset(x, y, sizeof(x))
#define SZ(x) ((int)(x).size())

using namespace std;

typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector<vi> vvi;
typedef long long i64;
typedef vector<i64> vi64;
typedef vector<vi64> vvi64;
typedef pair<i64, i64> pi64;
typedef double ld;

template<class T> bool uin(T &a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool uax(T &a, T b) { return a < b ? (a = b, true) : false; }

const int maxn = 2000+100;
int n, m, dp[maxn][maxn], val[30];
char s[maxn];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.precision(10);
    cout << fixed;
#ifdef LOCAL_DEFINE
    freopen("input.txt", "r", stdin);
#endif

    scanf("%d%d", &n, &m);
    scanf("%s", (s + 1));
    getchar();
    char op;
    memset(val, 0x3f, sizeof(val));
    for (int i = 0; i < n; ++i) {
        scanf("%c", &op);
        int x, y;
        scanf("%d%d", &x, &y);
        getchar();
        val[int(op - 'a')] = min(val[int(op - 'a')], min(x, y));
    }
    for (int i = 1; i <= m; ++i) dp[i][i] = 0;
    for (int len = 1; len <= m; ++len) {
        for (int i = 1; i + len - 1 <= m; ++i) {
            int j = i + len - 1;
            if (s[i] == s[j]) dp[i][j] = dp[i + 1][j - 1];
            else {
                dp[i][j] = min(dp[i + 1][j] + val[int(s[i] - 'a')], dp[i][j - 1] + val[int(s[j] - 'a')]);
            }
        }
    }
    cout << dp[1][m] << '\n';

#ifdef LOCAL_DEFINE
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
    return 0;
}