usaco3.3.5 A Game

本文介绍了一个两玩家博弈游戏,目标是在交替从两端移除并累加数字的过程中获得比对手更高的总分。通过动态规划方法实现了最优策略,确保了在面对最佳对手时也能获取最大利益。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

一 原题

A Game
IOI'96 - Day 1

Consider the following two-player game played with a sequence of N positive integers (2 <= N <= 100) laid onto a 1 x N game board. Player 1 starts the game. The players move alternately by selecting a number from either the left or the right end of the gameboar. That number is then deleted from the board, and its value is added to the score of the player who selected it. A player wins if his sum is greater than his opponents.

Write a program that implements the optimal strategy. The optimal strategy yields maximum points when playing against the "best possible" opponent. Your program must further implement an optimal strategy for player 2.

PROGRAM NAME: game1

INPUT FORMAT

Line 1:N, the size of the board
Line 2-etc:N integers in the range (1..200) that are the contents of the game board, from left to right

SAMPLE INPUT (file game1.in)

6
4 7 2 9
5 2

OUTPUT FORMAT

Two space-separated integers on a line: the score of Player 1 followed by the score of Player 2.

SAMPLE OUTPUT (file game1.out)

18 11


二 分析

动态规划。dp[i][j]表示在区间[i, j]内先手能获得的最高分数。


三 代码

运行结果:
USER: Qi Shen [maxkibb3]
TASK: game1
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.000 secs, 4264 KB]
   Test 2: TEST OK [0.000 secs, 4264 KB]
   Test 3: TEST OK [0.000 secs, 4264 KB]
   Test 4: TEST OK [0.000 secs, 4264 KB]
   Test 5: TEST OK [0.000 secs, 4264 KB]
   Test 6: TEST OK [0.000 secs, 4264 KB]
   Test 7: TEST OK [0.000 secs, 4264 KB]
   Test 8: TEST OK [0.000 secs, 4264 KB]
   Test 9: TEST OK [0.000 secs, 4264 KB]
   Test 10: TEST OK [0.000 secs, 4264 KB]
   Test 11: TEST OK [0.000 secs, 4264 KB]
   Test 12: TEST OK [0.000 secs, 4264 KB]
   Test 13: TEST OK [0.000 secs, 4264 KB]
   Test 14: TEST OK [0.000 secs, 4264 KB]
   Test 15: TEST OK [0.000 secs, 4264 KB]
   Test 16: TEST OK [0.000 secs, 4264 KB]

All tests OK.

Your program ('game1') produced all correct answers! This is your submission #2 for this problem. Congratulations!


AC代码:
/*
ID:maxkibb3
LANG:C++
PROG:game1
*/

#include<cstdio>
#include<algorithm>
using namespace std;

const int MAXN = 105;

int n;
int a[MAXN];
int sum[MAXN][MAXN];
int dp[MAXN][MAXN];

int main() {
    freopen("game1.in", "r", stdin);
    freopen("game1.out", "w", stdout);
    scanf("%d", &n);
    for(int i = 0; i < n; i++) {
        scanf("%d", &a[i]);
        sum[i][i] = a[i];
        dp[i][i] = a[i];
    }
    for(int i = 0; i < n; i++) {
        for(int j = i + 1; j < n; j++) {
            sum[i][j] = sum[i][j - 1];
            sum[i][j] += a[j];
        }
    }
    for(int i = 1; i < n; i++) {
        for(int j = 0; j < n - i; j++) {
            dp[j][j + i] = max(a[j] + sum[j + 1][j + i] - dp[j + 1][j + i],
                           a[j + i] + sum[j][j + i - 1] - dp[j][j + i - 1]);
        }
    }
    printf("%d %d\n", dp[0][n - 1], sum[0][n - 1] - dp[0][n - 1]);
    return 0;
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值