【BZOJ4380】【POI2015】Myjnie（区间DP）

Description

有n家洗车店从左往右排成一排，每家店都有一个正整数价格p[i]。有m个人要来消费，第i个人会驶过第a[i]个开始一直到第b[i]个洗车店，且会选择这些店中最便宜的一个进行一次消费。但是如果这个最便宜的价格大于c[i]，那么这个人就不洗车了。请给每家店指定一个价格，使得所有人花的钱的总和最大。

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Solution

设fl,r,k f l , r , k 表示只考虑完全被包含于[l,r] [ l , r ] 的区间、[l,r] [ l , r ] 的最小值为k k 时的最大利润。
那么枚举m∈[l,r]m∈[l,r]，有：fl,r,k=maxfl,m−1,x+fm+1,r,y+suml,r,m,kk f l , r , k = max f l , m − 1 , x + f m + 1 , r , y + s u m l , r , m , k k
其中x≥k,y≥k x ≥ k , y ≥ k ，suml,r,m,k s u m l , r , m , k 表示完全被包含于[l,r] [ l , r ] 的区间中跨过m m 且cc大于等于k k <script type="math/tex" id="MathJax-Element-576">k</script>的区间有多少个。
直接转移即可。

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Code

/************************************************
 * Au: Hany01
 * Date: Aug 28th, 2018
 * Prob: BZOJ4380 POI2015 Myjnie

 * Inst: Yali High School
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef long double LD;
typedef pair<int, int> PII;
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define SZ(a) ((int)(a).size())
#define ALL(a) a.begin(), a.end()
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read() {
    static int _, __; static char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 53, maxm = 4003;

int n, m, a[maxm], b[maxm], c[maxm], ls[maxm], stk[maxm], top, cur, g[maxn][maxn][maxm], f[maxn][maxn][maxm], fp[maxn][maxn][maxm], gp[maxn][maxn][maxm], tot;

void print(int l, int r, int k) {
    if (l > r) return;
    if (l == r) printf("%d ", ls[gp[l][r][k]]);
    else
        print(l, fp[l][r][gp[l][r][k]] - 1, gp[l][r][k]),
        printf("%d ", ls[gp[l][r][k]]),
        print(fp[l][r][gp[l][r][k]] + 1, r, gp[l][r][k]);
}

int main()
{
#ifdef hany01
    freopen("bzoj4380.in", "r", stdin);
    freopen("bzoj4380.out", "w", stdout);
#endif

    n = read(), m = read();
    For(i, 1, m) a[i] = read(), b[i] = read(), c[i] = ls[i] = read();
    sort(ls + 1, ls + 1 + m), tot = unique(ls + 1, ls + 1 + m) - ls - 1;
    For(i, 1, m) c[i] = lower_bound(ls + 1, ls + 1 + tot, c[i]) - ls;

    Fordown(l, n, 1) For(r, l, n) {
        For(k, 1, tot) fp[l][r][k] = l;
        For(p, l, r) {
            top = 0;
            For(i, 1, m) if (l <= a[i] && b[i] <= r && a[i] <= p && p <= b[i]) stk[++ top] = c[i];
            sort(stk + 1, stk + 1 + top), cur = 0;
            For(k, 1, tot) {
                while (cur < top && stk[cur + 1] < k) ++ cur;
                if (chkmax(f[l][r][k], g[l][p - 1][k] + g[p + 1][r][k] + (top - cur) * ls[k])) fp[l][r][k] = p;
            }
        }
        Fordown(k, tot, 1) {
            g[l][r][k] = f[l][r][k], gp[l][r][k] = k;
            if (chkmax(g[l][r][k], g[l][r][k + 1])) gp[l][r][k] = gp[l][r][k + 1];
        }
    }
    printf("%d\n", g[1][n][1]), print(1, n, 1);

    return 0;
}