c/c++语言开发共享noi.ac #289. 电梯(单调队列)

题意

sol

傻叉的我以为给出的(t)是单调递增的，然后(100rightarrow0)

首先可以按(t)排序，那么转移方程为

(f[i] = min_{j=0}^{i-1}(max(t[i], f[j]) + 2 * max_{k=j+1}^i x[k]))

不难发现，若(i < j)且(x[i] < x[j])，那么从(i)转移过来一定是不优的，一定是从(i)之前的某个位置转移过来。(f单增)

然后直接单调队列搞一搞就行了，

#include<bits/stdc++.h>  #define pair pair<int, int> #define fi first #define se second #define ll long long  #define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? eof : *p1++) char buf[(1 << 22)], *p1 = buf, *p2 = buf; using namespace std; const int maxn = 1e6 + 10; template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;} template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;} inline int read() {     char c = getchar(); int x = 0, f = 1;     while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}     while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();     return x * f; } int n, q[maxn], val[maxn], top; ll f[maxn]; pair a[maxn]; signed main() {     n = read();      for(int i = 1; i <= n; i++) a[i].fi = read(), a[i].se = read();     sort(a + 1, a + n + 1);     for(int i = 1; i <= n; i++) {         while(top && a[i].se > a[top].se) top--;         a[++top] = a[i];     }     memset(f, 0x7f, sizeof(f));     n = top; f[0] = 0;     int l = 1, r = 0, las = 0;     for(int i = 1; i <= n; i++) {         while(l <= r && a[i].fi >= f[q[l]]) las = q[l++];         if(las < i) chmin(f[i], a[i].fi + 2ll * a[las + 1].se);         if(l <= r) chmin(f[i], val[l]);         int cur = f[i] + 2ll * a[i + 1].se;         while(l <= r && cur <= val[r]) r--;         q[++r] = i; val[r] = cur;     }     cout << f[n];     return 0; }