Codeforces Round #189 (Div. 1) C

C - Kalila and Dimna in the Logging Industry

很容易能得到状态转移方程 dp[ i ] = min( dp[ j ] + b[ j ] * a[ i ] )， 然后斜率优化一下。

一直以为炸精度了， 忽然发现手贱把while 写成了if 。。。。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long
using namespace std;

const int N = 4e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;

int n, que[N], head, rear;
LL a[N], b[N], dp[N];

double calc(int k, int j) {
    return 1.0 * (dp[j] - dp[k]) * (b[k] - b[j]);
}

int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%lld", &a[i]);
    for(int i = 1; i <= n; i++) scanf("%lld", &b[i]);
    dp[1] = 0;
    head = 1, rear = 0;
    que[++rear] = 1;
    for(int i = 2; i <= n; i++) {
        while(rear - head + 1 >= 2 && calc(que[head], que[head + 1]) < a[i]) head++;
        dp[i] = dp[que[head]] + b[que[head]] * a[i];
        while(rear - head + 1 >= 2  && calc(que[rear-1], que[rear]) >= calc(que[rear], i)) rear--;
        que[++rear] = i;
    }
    printf("%lld
", dp[n]);
    return 0;
}

/*
*/