HDU-5074
Hatsune Miku
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 931 Accepted Submission(s): 651
Today you want to compose a song, which is just a sequence of notes. There are only m different notes provided in the package. And you want to make a song with n notes.
Also, you know that there is a system to evaluate the beautifulness of a song. For each two consecutive notes a and b, if b comes after a, then the beautifulness for these two notes is evaluated as score(a, b).
So the total beautifulness for a song consisting of notes a1, a2, . . . , an, is simply the sum of score(ai, ai+1) for 1 ≤ i ≤ n - 1.
Now, you find that at some positions, the notes have to be some specific ones, but at other positions you can decide what notes to use. You want to maximize your song’s beautifulness. What is the maximum beautifulness you can achieve?
For each test case, the first line contains two integers n(1 ≤ n ≤ 100) and m(1 ≤ m ≤ 50) as mentioned above. Then m lines follow, each of them consisting of m space-separated integers, the j-th integer in the i-th line for score(i, j)( 0 ≤ score(i, j) ≤ 100). The next line contains n integers, a1, a2, . . . , an (-1 ≤ ai ≤ m, ai ≠ 0), where positive integers stand for the notes you cannot change, while negative integers are what you can replace with arbitrary notes. The notes are named from 1 to m.
2
5 3
83 86 77
15 93 35
86 92 49
3 3 3 1 2
10 5
36 11 68 67 29
82 30 62 23 67
35 29 2 22 58
69 67 93 56 11
42 29 73 21 19
-1 -1 5 -1 4 -1 -1 -1 4 -1
/**
题意:m种点,一个长度为n的串,若串中的数为-1 则为待定,否则为这点
求这个串的最大价值,
做法:dp
分为四种情况,如果当前点是未知
前一个点已知
前一个点未知
如果当前点已知
前一个点已知
前一个点未知
情况很少 很容易枚举
============================================================
dp是硬伤
============================================================
**/
#include <iostream>
#include <algorithm>
#include <string.h>
#include <cmath>
#include <stdio.h>
#include <queue>
#define maxn 110
using namespace std;
int score[maxn][maxn];
int note[maxn];
int dp[maxn][maxn];
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
int n, m;
scanf("%d %d", &n, &m);
for(int i = ; i <= m; i++)
{
for(int j = ; j <= m; j++)
{
scanf("%d", &score[i][j]);
}
}
for(int i = ; i <= n; i++)
{
scanf("%d", ¬e[i]);
}
memset(dp, , sizeof(dp));
for(int i = ; i <= n; i++)
{
if(note[i] > )
{
if(note[i - ] > )
{
dp[i][note[i]] = dp[i - ][note[i - ]] + score[note[i - ]][note[i]];
}
else
{
for(int j = ; j <= m; j++)
{
dp[i][note[i]] = max(dp[i][note[i]], dp[i - ][j] + score[j][note[i]]);
}
}
}
else
{
if(note[i - ] > )
{
for(int j = ; j <= m; j++)
{
dp[i][j] = max(dp[i][j], dp[i - ][note[i - ]] + score[note[i - ]][j]);
}
}
else
{
for(int j = ; j <= m; j++)
{
for(int k = ; k <= m; k++)
{
dp[i][j] = max(dp[i][j], dp[i - ][k] + score[k][j]);
}
}
}
}
}
int sum = -;
if(note[n] > ) {
sum = dp[n][note[n]];
}
else
{
for(int i = ; i <= m; i++)
{
sum = max(sum, dp[n][i]);
}
}
printf("%d\n", sum);
}
return ;
}
HDU-5074的更多相关文章
- hdu 5074 Hatsune Miku DP题目
题目传送门http://acm.hdu.edu.cn/showproblem.php?pid=5074 $dp[i][j] =$ 表示数列前$i$个数以$j$结尾的最大分数 $dp[i][j] = - ...
- HDU 5074 Hatsune Miku(2014鞍山赛区现场赛E题)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5074 解题报告:给出一个长度为n的序列,例如a1,a2,a3,a4......an,然后这个序列的美丽 ...
- dp --- 2014 Asia AnShan Regional Contest --- HDU 5074 Hatsune Miku
Hatsune Miku Problem's Link: http://acm.hdu.edu.cn/showproblem.php?pid=5074 Mean: 有m种音符(note),现在要从 ...
- [HDU 5074] Hatsune Miku (动态规划)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5074 题目大意是给你m个note,n个数,得分是v[a[i]][a[i+1]]的总和,如果说a[i]是 ...
- hdu 5074 Hatsune Miku
http://acm.hdu.edu.cn/showproblem.php?pid=5074 题意:给你一个的矩阵score[i][j],然后给你一个数列,数列中有一些是-1,代表这个数可以换成1~m ...
- hdu 5074 相邻的和最大dp
http://acm.hdu.edu.cn/showproblem.php?pid=5074 给定一个序列 有些位数未知,给你全部两个数连续所得到的能量.问你怎么安排数字使得总能量最大 二维dp,dp ...
- hdu 5074 相邻数和最大dp
http://acm.hdu.edu.cn/showproblem.php?pid=5074 给定一个序列 有些位数未知,给你所有两个数连续所得到的能量,问你怎么安排数字使得总能量最大 二维dp,dp ...
- HDU 5074 Luck Competition (暴力,概率)
题意:有 n 个人参加比赛,给出n-1个人的成绩,然后要选出一个幸运的人,先把所有的分数求平均数,然后再*2/3,那个不大于这个数,且最接近的数,就是最幸运的, 让你设置最后一个人的分,使他是最幸运的 ...
- HDU 5074 Hatsune Miku(DP)
Problem Description Hatsune Miku is a popular virtual singer. It is very popular in both Japan and C ...
- dp HDU - 5074
按题意推表达式 #include<cstdio> #include<cstring> #define max(a, b) (a)>(b)?(a):(b) ][], num ...
随机推荐
- HDU 4010 Query on The Trees(动态树LCT)
Problem Description We have met so many problems on the tree, so today we will have a query problem ...
- JavaSE复习(八)反射和注解
反射 框架设计的灵魂 框架:半成品软件.可以在框架的基础上进行软件开发,简化编码 反射:将类的各个组成部分封装为其他对象,这就是反射机制 好处: 可以在程序运行过程中,操作这些对象. 可以解耦,提高程 ...
- 揭开网络编程常见API的面纱【上】
Linux网络编程API函数初步剖析 今天我们来分析一下前几篇博文中提到的网络编程中几个核心的API,探究一下当我们调用每个API时,内核中具体做了哪些准备和初始化工作. 1.socket(famil ...
- 团队作业4——第一次项目冲刺(Alpha版本)-第二篇
项目冲刺——第二阶段 度过了敏捷冲刺,各个成员积极汇报各自的工作.好了,着手下一步规划! Mission 团队成员 任务 郭达 实现PHP后台的答题判分查看正确率 刘德培 编写博客 石浩洋 实现 ...
- lintcode-78-最长公共前缀
78-最长公共前缀 给k个字符串,求出他们的最长公共前缀(LCP) 样例 在 "ABCD" "ABEF" 和 "ACEF" 中, LCP 为 ...
- HDU 2139 Calculate the formula
http://acm.hdu.edu.cn/showproblem.php?pid=2139 Problem Description You just need to calculate the su ...
- 【SSH】——spring的控制反转和依赖注入
spring是一个轻量级的容器框架,主要是为了使企业的开发变得简单.高效.无论是从大小还是开销来讲,他都可以算是轻量级的,也是非侵入性的. 下图是spring的框架示意图,说到spring,就不得不提 ...
- Linux挂载Win共享文件夹_VmwareTools
- 【PHP】- Apache设置
Apache配置 1.首先新建一个自己的amp目录(模仿wampserver安装目录),以后的apache,mysql,php都放在此目录下. 2.下载apache 根据自己的系统下载相应的压缩包,我 ...
- Log4Net讲解
声明:本文内容主要译自Nauman Leghari的Using log4net,亦加入了个人的一点心得(节3.1.4). 1 简介 1.1 Log4net的优点: ...