随笔 - 375  文章 - 2  评论 - 104  0

蒟蒻就写一下简单题吧。。。

此题很容易写出dp的方程:

令f[i]表示i号点要建仓库的话最小总费用,则

f[i] = min(f[j]  + (x[i] - x[j + 1]) * p[i + 1] + (x[i] - x[j + 2])* p[i + 2] + ... + (x[i] - x[i]) * p[i]) + c[i]

然后把min里面的东西展开再合并,并且令

sp[i] = p[1] + p[2] + ... + p[i]

s[i] = x[1] * p[1] + x[2] * p[2] + ... + x[i] * p[i]

那么:

f[i] = min(f[j] + s[j] - x[i] * sp[j]) + c[i] + sp[i] * x[i] - s[i];

然后前面的min里的式子可以斜率优化掉,就做完了。(我好像会自己推了?进步+1^_^)

 

 

 1 #include <cstdlib>
 2 #include <cstdio>
 3 #include <iostream>
 4  
 5 using namespace std;
 6  
 7 long long q[1200000], s[1200000], sp[1200000], f[1200000], g[1200000];
 8 long long x;
 9 int l = 1, r = 1, n;
10  
11 inline bool pop_head(){
12     int a = q[l], b = q[l + 1];
13     return (long long) g[b] - g[a] < (long long)x * (sp[b] - sp[a]);
14 }
15  
16 inline bool pop_tail(int i){
17     int a = q[r - 1], b = q[r];
18     return (long long) (g[b] - g[a]) * (sp[i] - sp[b]) >= (long long) (g[i] - g[b]) * (sp[b] - sp[a]);
19 }
20  
21 int main(){
22     scanf("%d\n", &n);
23     long long p, c;
24     q[1] = 0;
25     int j;
26     for (int i = 1; i <= n; ++i){
27         scanf("%lld %lld %lld\n", &x, &p, &c);
28         sp[i] = sp[i - 1] + p;
29         s[i] = s[i - 1] + x * p;
30         while (l < r && pop_head()) ++l;
31         j = q[l];
32         f[i] = g[j] - x * sp[j] + c + sp[i] * x - s[i];
33         g[i] = f[i] + s[i];
34         while (l < r && pop_tail(i)) --r;
35         q[++r] = i;
36     }
37     printf("%lld\n", f[n]);
38     return 0;
39 }
View Code

(p.s. 沙茶的我把q写成g,结果WA了两次。。。)

posted on 2014-10-11 15:54  Xs酱~  阅读(111)  评论(0编辑  收藏