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Sicily 1176. Two Ends

2010-07-26 20:50:02| 分类：动态规划 | 标签： |举报 |字号大中小订阅

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      题目链接： http://www.sicily.3322.org/sicily/show_problem.php?pid=1176

      动态规划。
      设 d [ a ][ b ] 表示在区间 [ a, b ] 上，玩家 2 运用贪心策略所导致双方分数的最大差值。
      递推公式：
      由于玩家 1 可以任意选择在两头取数，因此要对玩家 1 每次的选择进行分类讨论。

      1. 玩家 1 选择取最左边的数 array[ a ]
      if (array[a + 1] >= array[b])
      points1 = dp(a + 2, b) + array[a] - array[a + 1]    (玩家 2 选择array[a + 1])
      else
      points1 = dp(a + 1, b - 1) + array[a] - array[b]     (玩家 2 选择array[b])

      2. 玩家 2 选择取最右边的数 array[ b ]
      if (array[a] >= array[b - 1])
      points2 = dp(a + 1, b - 1) + array[b] - array[a]     (玩家 2 选择array[a])
      else
      points2 = dp(a, b - 2) + array[b] - array[b - 1]      (玩家 2 选择array[b - 1])

      那么，
      d[a][b] = max(points1, points2)

      代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int oo = 100100;

int array[1100];
int d[1100][1100];

int dp(int a, int b)
{
    if (d[a][b] != oo)
    return d[a][b];

    if (a > b)
    return 0;

    if (array[a + 1] >= array[b] && array[a] >= array[b - 1])
    d[a][b] = max(dp(a + 2, b) + array[a] - array[a + 1], dp(a + 1, b - 1) + array[b] - array[a]);
    else if (array[a + 1] >= array[b] && array[a] < array[b - 1])
    d[a][b] = max(dp(a + 2, b) + array[a] - array[a + 1], dp(a, b - 2) + array[b] - array[b - 1]);
    else if (array[a + 1] < array[b] && array[a] >= array[b - 1])
    d[a][b] = max(dp(a + 1, b - 1) + array[a] - array[b], dp(a + 1, b - 1) + array[b] - array[a]);
    else
    d[a][b] = max(dp(a + 1, b - 1) + array[a] - array[b], dp(a, b - 2) + array[b] - array[b - 1]);

    return d[a][b];
}

int main()
{
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);

    int n;
    int counter;

    counter = 0;
    while (scanf("%d", &n), n != 0)
    {
        counter ++;
        for (int i = 1; i <= n; i ++)
        for (int j = 1; j <= n; j ++)
        d[i][j] = oo;

        for (int i = 1; i <= n; i ++)
        scanf("%d", &array[i]);

        printf("In game %d, the greedy strategy might lose by as many as %d points.\n", counter, dp(1, n));
    }

    return 0;
}

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