UVA 10891 Game of Sum(DP)
This is a two player game. Initially there are n integer numbers in an array and players A and B get chance to take them alternatively. Each player can take one or more numbers from the left or right end of the array but cannot take from both ends at a time. He can take as many consecutive numbers as he wants during his time. The game ends when all numbers are taken from the array by the players. The point of each player is calculated by the summation of the numbers, which he has taken. Each player tries to achieve more points from other. If both players play optimally and player A starts the game then how much more point can player A get than player B?
Input
The input consists of a number of cases. Each case starts with a line specifying the integer n (0 < n ≤100), the number of elements in the array. After that, nnumbers are given for the game. Input is terminated by a line where n=0.
Output
For each test case, print a number, which represents the maximum difference that the first player obtained after playing this game optimally.
题目大意:给n个数,两个人轮流取数,可以从左往右或从右往左取任意多个。两个人都希望自己的取得的数的总和尽量大,都采取最优策略,问第一个人能比第二个人取得的数多多少。
思路:很容易可以想到一个$O(n^3)$的DP,用dp[i][j]代表只剩下a[i..j]的数,先手可以取得的最大值,此时后手取得的最大值为sum[i..j] - dp[i][j]。
那么状态转移方程为:dp[i][j] = max(sum[i..j], sum[i..j] - min(dp[i+1][j], dp[i+2][j]……), sum[i..j] - min(dp[i][j - 1], dp[i, j - 2])。
输出结果为2 * dp[1][n] - sum[1..n]。
代码(0.026S):
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std; const int MAXN = ; int dp[MAXN][MAXN];
int a[MAXN], sum[MAXN];
int n; int main() {
while(scanf("%d", &n) != EOF && n) {
for(int i = ; i <= n; ++i) scanf("%d", a + i);
for(int i = ; i <= n; ++i) sum[i] = sum[i - ] + a[i];
for(int k = ; k < n; ++k) {
for(int i = ; i + k <= n; ++i) {
int j = i + k;
dp[i][j] = sum[j] - sum[i - ];
for(int p = i + ; p <= j; ++p) dp[i][j] = max(dp[i][j], sum[j] - sum[i - ] - dp[p][j]);
for(int p = j - ; p >= i; --p) dp[i][j] = max(dp[i][j], sum[j] - sum[i - ] - dp[i][p]);
}
}
printf("%d\n", * dp[][n] - sum[n]);
}
}
这个DP还有优化的余地,观察状态转移方程可以发现,dp[i][j]使用了min(dp[i+1][j], dp[i+2][j]……),而dp[i+1][j]=min(dp[i+2][j], dp[i+3][j]……),有重复的部分。
于是我们可以用l[i][j]记录max(dp[i][j], dp[i+1][j], dp[i+2][j]……),即从左往右取的后手最小值,则sum[i..j] - min(dp[i+1][j], dp[i+2][j]……)可以写成sum[i..j]-l[i+1][j]。每次更新l[i][j] = min(dp[i][j], l[i+1][j])。
同理用r[i][j]记录从右往左取的后手最小值。
至此DP优化至$O(n^2)$。
代码(0.015S):
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std; const int MAXN = ; int dp[MAXN][MAXN];
int l[MAXN][MAXN], r[MAXN][MAXN];
int a[MAXN], sum[MAXN];
int n; int main() {
while(scanf("%d", &n) != EOF && n) {
for(int i = ; i <= n; ++i) scanf("%d", a + i);
for(int i = ; i <= n; ++i) sum[i] = sum[i - ] + a[i];
for(int k = ; k < n; ++k) {
for(int i = ; i + k <= n; ++i) {
int j = i + k;
l[i][j] = r[i][j] = dp[i][j] = sum[j] - sum[i - ];
if(i != j) {
dp[i][j] = max(dp[i][j], sum[j] - sum[i - ] - l[i + ][j]);
dp[i][j] = max(dp[i][j], sum[j] - sum[i - ] - r[i][j - ]);
l[i][j] = min(dp[i][j], l[i + ][j]);
r[i][j] = min(dp[i][j], r[i][j - ]);
}
}
}
printf("%d\n", * dp[][n] - sum[n]);
}
}
UVA 10891 Game of Sum(DP)的更多相关文章
- uva 10891 Game of Sum(区间dp)
题目连接:10891 - Game of Sum 题目大意:有n个数字排成一条直线,然后有两个小伙伴来玩游戏, 每个小伙伴每次可以从两端(左或右)中的任意一端取走一个或若干个数(获得价值为取走数之和) ...
- UVA 10891 Game of Sum(区间DP(记忆化搜索))
题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem ...
- UVA - 10891 Game of Sum (区间dp)
题意:AB两人分别拿一列n个数字,只能从左端或右端拿,不能同时从两端拿,可拿一个或多个,问在两人尽可能多拿的情况下,A最多比B多拿多少. 分析: 1.枚举先手拿的分界线,要么从左端拿,要么从右端拿,比 ...
- Max Sum (dp)
Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. F ...
- URAL 1146 Maximum Sum(DP)
Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the large ...
- UVA - 10891 Game of Sum 区间DP
题目连接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=19461 Game of sum Description This ...
- HDU 1003 Max Sum(DP)
点我看题目 题意 : 就是让你从一个数列中找连续的数字要求他们的和最大. 思路 : 往前加然后再判断一下就行. #include <iostream> #include<stdio. ...
- 【UVa】Partitioning by Palindromes(dp)
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=27&page=sh ...
- 【noi 2.6_1481】Maximum sum(DP)
题意:求不重叠的2段连续和的最大值. 状态定义f[i]为必选a[i]的最大连续和,mxu[i],mxv[i]分别为前缀和后缀的最大连续和. 注意:初始化f[]为0,而max值为-INF.要看好数据范围 ...
随机推荐
- nrf51822裸机教程-GPIO
首先看看一下相关的寄存器说明 Out寄存器 输出设置寄存器 每个比特按顺序对应每个引脚,bit0对应的就是 引脚0 该寄存器用来设置 引脚作为输出的时候的 输出电平为高还是低. 与输出设置相关的 还有 ...
- JavaScript POST 请求如何跨域
前天遇到一个问题,就是我上传图片的时候,这个图片需要上传给某个API的接口 这样问题就来了,我们之前上传图片的时候都是先上传到我们自己的后台里面,然后通过后台,再把这个流传到图片服务器上. 但是上传到 ...
- iOS开发教程之:iPhone开发环境搭建
安装条件: 硬件:一台拥有支持虚拟技术的64位双核处理器和2GB以上内存的PC. 注意:运行MAC OS,需要电脑支持虚拟技术(VT),安装时,需要将VT启动,在BIOS中开启. 软件: Window ...
- [LeetCode]题解(python):038-Count and Say
题目来源 https://leetcode.com/problems/count-and-say/ The count-and-say sequence is the sequence of inte ...
- sublime text2 操作及插件
sublime text2 1. 文件快速导航: 这是sublime上面很好用的功能之一,ctrl+p可以调出窗口,菜单上的解释是gotoanythings ,确实如其所言,调出窗口后,直接输入关键字 ...
- c# 并行运算
c# 并行运算 1. Parallel.INVOKE() 看实例: private static Stopwatch watch = new Stopwatch(); private static v ...
- Aspose.word在asp.net mvc中如何使用的个人总结
项目需要导出数据到word中,因为要导出的是表格形式,所以先在word中绘制好了表格,然后按照以前的代码改了改,发现不行.出现的问题如下: 这是当时的代码,问题的症结所在就是Response上.这段代 ...
- tabBaritem的图片偏移
tabBarItem.imageInsets = UIEdgeInsetsMake(-10, 0, 10, 0);
- 求以下表达式的值,写出您想到的一种或几种实现方法: 1-2+3-4+……+m
private static int fun(int m) { ; ; i <= m; i++) { == ) temp = temp + i; else temp = temp - i; } ...
- sdut2169:Sequence(dp)
题目:http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2169 #include <iost ...