【ZJOI2017 Round1练习】D8T2 sequence（DP）

题意：

思路：

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <map>
using namespace std;

const int N = 153, INF = -0x3f3f3f3f;
int n, m, ans[N], v[N], a[N], last[N][N], f[N][N][N], g[N][N][N];
map<int, int> hash;

inline void upt(int &x, const int &y) {
    if (x < y) x = y;
}

char ch;
inline int read() {
    int res = 0, sgn = 0;
    while (ch = getchar(), ch < '0' || ch > '9') if (ch == '-') break;
    ch == '-' ? sgn = 1 : res = ch - 48;
    while (ch = getchar(), ch >= '0' && ch <= '9') res = res * 10 + ch - 48;
    return sgn == 0 ? res : -res;
}

int main() {
  freopen("sequence.in", "r", stdin);
  freopen("sequence.out", "w", stdout);

    n = read();
    for (int i = 1; i <= n; ++i) v[i] = read();
    for (int i = 1; i <= n; ++i) {
        a[i] = read();
        if (hash[a[i]] == 0) hash[a[i]] = ++m;
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) last[i][j] = last[i - 1][j];
        last[i][hash[a[i]]] = i;
    }
    memset(f, INF, sizeof(f));
    memset(g, INF, sizeof(g));
    for (int i = 1; i <= n; ++i) {
        for (int j = 0; j <= n; ++j)
            if (i + j <= n)
                f[i][i][j] = g[i][i][j] = v[j + 1];
            else     break;
        f[i][i - 1][0] = g[i][i - 1][0] = 0;
    }
    for (int l = 1; l <= n; ++l)
        for (int i = 1; i <= n; ++i) {
            int j = i + l;
            if (j > n) break;
            for (int k = 0; k <= n; ++k) {
                if (j + k > n) break;
                upt(f[i][j][k], f[i][j - 1][0] + v[k + 1]);
                upt(g[i][j][k], f[i][j - 1][0] + v[k + 1]);
                int p = hash[a[j] + 1], q = hash[a[j] - 1];
                for (int h = last[j][p]; h >= i; h = last[h - 1][p])
                    upt(f[i][j][k], f[i][h][k + 1] + f[h + 1][j - 1][0]);
                for (int h = last[j][q]; h >= i; h = last[h - 1][q]) {
                    upt(f[i][j][k], g[i][h][k + 1] + f[h + 1][j - 1][0]);
                    upt(g[i][j][k], g[i][h][k + 1] + f[h + 1][j - 1][0]);
                }
            }
        }
    ans[0] = 0;
    for (int i = 1; i <= n; ++i) {
        upt(ans[i], ans[i - 1]);
        for (int j = 0; j < i; ++j)
            upt(ans[i], ans[j] + f[j + 1][i][0]);
    }
    printf("%d\n", ans[n]);
    return 0;
}