• 【HDOJ】3480 Division


    斜率dp+滚动数组。

      1 /* 3480 */
      2 #include <iostream>
      3 #include <sstream>
      4 #include <string>
      5 #include <map>
      6 #include <queue>
      7 #include <set>
      8 #include <stack>
      9 #include <vector>
     10 #include <deque>
     11 #include <algorithm>
     12 #include <cstdio>
     13 #include <cmath>
     14 #include <ctime>
     15 #include <cstring>
     16 #include <climits>
     17 #include <cctype>
     18 #include <cassert>
     19 #include <functional>
     20 #include <iterator>
     21 #include <iomanip>
     22 using namespace std;
     23 //#pragma comment(linker,"/STACK:102400000,1024000")
     24 
     25 #define sti                set<int>
     26 #define stpii            set<pair<int, int> >
     27 #define mpii            map<int,int>
     28 #define vi                vector<int>
     29 #define pii                pair<int,int>
     30 #define vpii            vector<pair<int,int> >
     31 #define rep(i, a, n)     for (int i=a;i<n;++i)
     32 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
     33 #define clr                clear
     34 #define pb                 push_back
     35 #define mp                 make_pair
     36 #define fir                first
     37 #define sec                second
     38 #define all(x)             (x).begin(),(x).end()
     39 #define SZ(x)             ((int)(x).size())
     40 #define lson            l, mid, rt<<1
     41 #define rson            mid+1, r, rt<<1|1
     42 
     43 const int maxn = 10005;
     44 const int maxm = 5005;
     45 int dp[2][maxn];
     46 int Q[maxn];
     47 int a[maxn];
     48 int n, m;
     49 
     50 void solve() {
     51     int l, r;
     52     int p = 0, q = 1;
     53     
     54     rep(i, 1, n+1)
     55         dp[p][i] = (a[i] - a[1]) * (a[i] - a[1]);
     56     
     57     rep(i, 2, m+1) {
     58         l = r = 0;
     59         Q[r++] = i-1;
     60         rep(j, i, n+1) {
     61             while (l+1 < r) {
     62                 int k1 = Q[l];
     63                 int k2 = Q[l+1];
     64                 int x1 = a[k1+1];
     65                 int x2 = a[k2+1];
     66                 int y1 = dp[p][k1] + x1 * x1;
     67                 int y2 = dp[p][k2] + x2 * x2;
     68                 if ((y2-y1)<=2*a[j]*(x2-x1))
     69                     ++l;
     70                 else
     71                     break;
     72             }
     73             int k = Q[l];
     74             int x = a[k+1];
     75             dp[q][j] = dp[p][k] + (a[j]-x)*(a[j]-x);
     76             while (l+1 < r) {
     77                 int k1 = j;
     78                 int k2 = Q[r-1];
     79                 int k3 = Q[r-2];
     80                 int x1 = a[k1+1];
     81                 int x2 = a[k2+1];
     82                 int x3 = a[k3+1];
     83                 int y1 = dp[p][k1] + x1 * x1;
     84                 int y2 = dp[p][k2] + x2 * x2;
     85                 int y3 = dp[p][k3] + x3 * x3;
     86                 if ((y3-y2)*(x2-x1) >= (y2-y1)*(x3-x2))
     87                     --r;
     88                 else
     89                     break;
     90             }
     91             Q[r++] = j;
     92             #ifndef ONLINE_JUDGE
     93                 printf("l = %d, r = %d
    ", l, r);
     94             #endif
     95         }
     96         q = p;
     97         p ^= 1;
     98     }
     99     
    100     int ans = dp[p][n];
    101     printf("%d
    ", ans);
    102 }
    103 
    104 int main() {
    105     ios::sync_with_stdio(false);
    106     #ifndef ONLINE_JUDGE
    107         freopen("data.in", "r", stdin);
    108         freopen("data.out", "w", stdout);
    109     #endif
    110     
    111     int t;
    112     
    113     scanf("%d", &t);
    114     rep(tt, 1, t+1) {
    115         scanf("%d %d", &n, &m);
    116         rep(i, 1, n+1)
    117             scanf("%d", &a[i]);
    118         sort(a+1, a+1+n);
    119         printf("Case %d: ", tt);
    120         solve();
    121     }
    122     
    123     #ifndef ONLINE_JUDGE
    124         printf("time = %d.
    ", (int)clock());
    125     #endif
    126     
    127     return 0;
    128 }
  • 相关阅读:
    02:AWT介绍
    01:GUI编程简介
    业余草 SpringCloud教程 | 第六篇: 分布式配置中心(Spring Cloud Config)(Finchley版本)
    业余草 SpringCloud教程 | 第五篇: 路由网关(zuul)(Finchley版本)
    业余草 SpringCloud教程 | 第四篇:断路器(Hystrix)(Finchley版本)
    业余草 SpringCloud教程 | 第三篇: 服务消费者(Feign)(Finchley版本)
    业余草 SpringCloud教程 | 第二篇: 服务消费者(rest+ribbon)(Finchley版本)
    业余草 SpringCloud 教程 | 第一篇: 服务的注册与发现Eureka(Finchley版本)
    业余草分享2018最新面试题总结
    业余草分享面试题,JVM结构、GC工作机制详解
  • 原文地址:https://www.cnblogs.com/bombe1013/p/5105950.html
Copyright © 2020-2023  润新知