HDU 4283 You Are the One(区间DP)
来源:互联网 发布:上证综合指数月度数据 编辑:程序博客网 时间:2024/05/29 02:16
You Are the One
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4622 Accepted Submission(s): 2192
Problem Description
The TV shows such as You Are the One has been very popular. In order to meet the need of boys who are still single, TJUT hold the show itself. The show is hold in the Small hall, so it attract a lot of boys and girls. Now there are n boys enrolling in. At the beginning, the n boys stand in a row and go to the stage one by one. However, the director suddenly knows that very boy has a value of diaosi D, if the boy is k-th one go to the stage, the unhappiness of him will be (k-1)*D, because he has to wait for (k-1) people. Luckily, there is a dark room in the Small hall, so the director can put the boy into the dark room temporarily and let the boys behind his go to stage before him. For the dark room is very narrow, the boy who first get into dark room has to leave last. The director wants to change the order of boys by the dark room, so the summary of unhappiness will be least. Can you help him?
Input
The first line contains a single integer T, the number of test cases. For each case, the first line is n (0 < n <= 100)
The next n line are n integer D1-Dn means the value of diaosi of boys (0 <= Di <= 100)
The next n line are n integer D1-Dn means the value of diaosi of boys (0 <= Di <= 100)
Output
For each test case, output the least summary of unhappiness .
Sample Input
2 512345554322
Sample Output
Case #1: 20Case #2: 24
Source
2012 ACM/ICPC Asia Regional Tianjin Online
周神的题解
#include <stdio.h>#include <iostream>#include <vector>#include <algorithm>#include <queue>#include <string.h>using namespace std;#define LL long longconst int maxn = 111;const int inf = 0x3f3f3f3f;int dp[maxn][maxn];int a[maxn];int sum[maxn];int main(){ int T; int p=0; scanf("%d",&T); while(T--){ int n;scanf("%d",&n); for(int i=1;i<=n;i++){ scanf("%d",&a[i]); sum[i]=sum[i-1]+a[i]; } for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ if(j<=i) dp[i][j]=0; else dp[i][j]=inf; } } for(int len=2;len<=n;len++){ for(int i=1;i<=n-len+1;i++){ int j=i+len-1; for(int k=i;k<=j;k++){ dp[i][j]=min(dp[i][j],dp[i][k-1]+dp[k+1][j]+(len-1)*a[k]+(k-i)*(sum[j]-sum[k])); } } } printf("Case #%d: ",++p); printf("%d\n",dp[1][n]); } return 0;}
阅读全文
0 0
- hdu 4283 You Are the One(区间dp)
- HDU 4283 You Are the One (区间DP)
- Hdu 4283 You Are the One(区间dp)
- HDU 4283 You are the one(区间DP)
- HDU 4283 You Are the One(区间DP)
- HDU 4283 You Are the One (区间dp)
- HDU 4283 You Are the One(区间DP)
- HDU 4283 You Are the One(区间DP)
- HDU 4283 You Are the One(区间DP)
- hdu 4283 You Are the One(区间DP)
- 【HDU 4283】You Are the One(区间DP)
- HDU 4283 You Are the One(区间DP)
- hdu 4283 You Are the One(区间dp)
- HDU 4283 You Are the One(区间dp)
- HDU 4283 You Are the One (区间dp)
- HDU 4283:You Are the One(区间DP)
- HDU 4283 You Are the One(区间DP)
- HDU 4283 You Are the One(区间DP)
- C++中关于输出精度与取整函数的问题
- ajax是什么? ajax的交互模型? 同步和异步的区别? 如何解决跨域问题?
- ZOJ 3993(2017CCPC秦皇岛站M题)Safest Buildings
- 简单计算器
- SurfaceView 实现动态曲线
- HDU 4283 You Are the One(区间DP)
- 1397 Goldbach's Conjecture
- c语言 rand()和srand()函数用法
- ubuntu 升级后无法连接wifi问题(Wireless 8265 / 8275网卡)
- 中文与url编码转换
- java-12-thread
- Android 搜索不到蓝牙设备
- How Big Is A Petabyte, Exabyte, Zettabyte, Or A Yottabyte?(COPY)
- PTA| 根据后序和中序遍历输出先序遍历(25 分)