有$n$间物品，每个物品有两个属性$W_i,H_i$，宽度跟高度，要求把这$n$件物品划分成若干连续的组，每组内$\sum W_i\le L$，并且要求最小化每组最大高度之和。

设$f[i]$表示以$i$为结尾的代价，则有：
$$f[i]=min(f[j]+ma{x}_{j\le k\le i}h[k])[\sum _{k=j}^{i}w[i]\le L]$$
容易发现这是可以$O(n^2)$转移的，考虑如何优化，

对于某个$f[i]$来说，以其$h[i]$作为最大值最多可以向左拓展的点，我们可以单调栈求出来，再利用二分查找，我们可以从$i$点向左拓展出一个点，满足$\sum w[i]\le L$.

之后我们只要在区间上查找最小值，作为当前点的答案，更新即可，

#include <bits/stdc++.h>
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;const int N = 1e6 + 10;long long ans[N], sum[N], minn[N << 2], mans[N << 2], lazy[N << 2], f[N];int n, m, h[N], w[N], stk[N], pre[N], top;void push_down(int rt) {if (lazy[rt]) {mans[ls] = minn[ls] + lazy[rt];mans[rs] = minn[rs] + lazy[rt];lazy[ls] = lazy[rs] = lazy[rt];lazy[rt] = 0;}
}void push_up(int rt) {minn[rt] = min(minn[ls], minn[rs]);mans[rt] = min(mans[ls], mans[rs]);
}void update(int rt, int l, int r, int x) {if (l == r) {mans[rt] = 0x3f3f3f3f3f3f3f3f, minn[rt] = f[x - 1];return ;}push_down(rt);if (x <= mid) {update(lson, x);}else {update(rson, x);}push_up(rt);
}void update(int rt, int l, int r, int L, int R, int x) {if (l >= L && r <= R) {lazy[rt] = x, mans[rt] = minn[rt] + x;return ;}push_down(rt);if (L <= mid) {update(lson, L, R, x);}if (R > mid) {update(rson, L, R, x);}push_up(rt);
}long long query(int rt, int l, int r, int L, int R) {if (l >= L && r <= R) {return mans[rt];}push_down(rt);long long ans = 0x3f3f3f3f3f3f3f3f;if (L <= mid) {ans = min(ans, query(lson, L, R));}if (R > mid) {ans = min(ans, query(rson, L, R));}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);scanf("%d %d", &n, &m);for (int i = 1; i <= n; i++) {scanf("%d %d", &h[i], &w[i]);sum[i] = sum[i - 1] + w[i];}stk[++top] = 1;for (int i = 2; i <= n; i++) {while (top && h[i] > h[stk[top]]) {top--;}pre[i] = stk[top], stk[++top] = i;}for (int i = 1; i <= n; i++) {update(1, 1, n, i);update(1, 1, n, pre[i] + 1, i, h[i]);int l = lower_bound(sum, sum + 1 + n, sum[i] - m) - sum;f[i] = query(1, 1, n, l + 1, i);}printf("%lld\n", f[n]);return 0;
}