题意
给定(m)个区间([a_i,b_i])以及(c_i),对于一个含有(n)个元素的序列(ans[]),区间(i)对其的贡献为(min{ans_i}(iin[a_i,b_i])<=c_i ? min{ans_i}(iin[a_i,b_i]) : 0),要求构造一个序列(ans[]),最大化区间的贡献之和。
(nleq50,mleq4000)
思路
离散化+区间( exttt{DP})
打死都不可能想到状态设计DP系列
稍作分析半天就会发现:存在一组答案使得每个(ans_i)都是某个(c_i)。因为把某个答案替换成第一个大于等于它的(c_i)不会更劣,因此(c_i)的值并不影响做题,但是大小顺序是有用的所以我们将(c_i)离散化。
因为一个区间的代价之和只与最小值有关,而且数据范围的(n)也不大,所以考虑区间( exttt{DP}):
设(f[l][r][k])表示区间([l,r])内(ans[])的最小值等于(k)的最大收益,(g[p][j])为当前区间穿过(p),且(cgeq j)的区间数量
枚举最小的位置(p),那么包含(p)的区间的答案全都是(k),之后转移
[f[l][r][k]=max(max(f[l][p - 1][k] + f[p + 1][r][k]+g[p][k]*k,pin[l,r]),f[l][r][k+1])
]
( exttt{DP})时顺便记录记录决策点,然后(dfs)输出
时间复杂度(O(n^3m))
代码
/*
Author:Loceaner
区间DP
*/
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int A = 51;
const int B = 4011;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
inline int read() {
char c = getchar(); int x = 0, f = 1;
for ( ; !isdigit(c); c = getchar()) if (c == '-') f = -1;
for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
return x * f;
}
struct node { int l, r, c; } a[B];
int n, m, tot, ans[B], res[B];
int pre[A][A][B], f[A][A][B], pos[A][A][B], g[A][B];
inline void work(int l, int r, int now) {
if (l > r) return;
int qwq = pos[l][r][now = pre[l][r][now]];
ans[qwq] = res[now];
work(l, qwq - 1, now), work(qwq + 1, r, now);
}
inline void update(int l, int r) {
for (int i = l; i <= r; i++)
for (int minn = 0; minn <= tot; minn++) g[i][minn] = 0;
for (int i = 1; i <= m; i++)
if (l <= a[i].l && a[i].r <= r)
for (int j = a[i].l; j <= a[i].r; j++) g[j][a[i].c]++;
for (int i = l; i <= r; i++)
for (int j = tot - 1; j >= 1; j--) g[i][j] += g[i][j + 1];
}
inline void dp(int l, int r) {
for (int k = tot; k >= 1; k--) {
int maxn = 0;
for (int p = l; p <= r; p++) {
int now = f[l][p - 1][k] + f[p + 1][r][k] + g[p][k] * res[k];
if (maxn <= now) maxn = now, pos[l][r][k] = p;
}
if (maxn >= f[l][r][k + 1]) f[l][r][k] = maxn, pre[l][r][k] = k;
else f[l][r][k] = f[l][r][k + 1], pre[l][r][k] = pre[l][r][k + 1];
}
}
signed main() {
n = read(), m = read();
for (int i = 1; i <= m; i++)
a[i].l = read(), a[i].r = read(), a[i].c = read(), res[i] = a[i].c;
sort(res + 1, res + 1 + m);
tot = unique(res + 1, res + m + 1) - res - 1;
for (int i = 1; i <= m; i++)
a[i].c = lower_bound(res + 1, res + tot + 1, a[i].c) - res;
for (int i = n; i >= 1; i--)
for (int j = i; j <= n; j++)
update(i, j), dp(i, j);
work(1, n, 1);
cout << f[1][n][1] << '
';
for (int i = 1; i <= n; i++) cout << ans[i] << " ";
return 0;
}