POJ1390：方盒游戏，考点：动态规划变量设置

原题：http://poj.org/problem?id=1390

描述

（由于百练上没有这道题，但是这道题又很重要，我印象中老师讲了好一会儿，所以还是写一写，题目描述直接照搬讲义了）

 

 

 

 

 问题分析

 

 

 解法

 

 

 

 

 

 

 

 

 

 代码如下：

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int M = 210;
struct Segment {
    int color;
    int len;
};
Segment segments[M];
int score[M][M][M];
int ClickBox(int i, int j, int len) {
    if (score[i][j][len] != -1)
        return score[i][j][len];
    int result = (segments[j].len + len) *
        (segments[j].len + len);
    if (i == j)
        return result;
    result += ClickBox(i, j - 1, 0);
    for (int k = i; k <= j - 1; ++k) {
        if (segments[k].color != segments[j].color)
            continue;
        int r = ClickBox(k + 1, j - 1, 0);
        r += ClickBox(i, k, segments[j].len + len);
        result = max(result, r);
    }
    score[i][j][len] = result;
    return result;
}
int main()
{
    int T;
    cin >> T;
    for (int t = 1; t <= T; ++t) {
        int n;
        memset(score, 0xff, sizeof(score));
        cin >> n;
        int lastC = 0;
        int segNum = -1;
        for (int i = 0; i < n; ++i) {
            int c;
            cin >> c;
            if (c != lastC) {
                segNum++;
                segments[segNum].len = 1;
                segments[segNum].color = c;
                lastC = c;
            }
            else segments[segNum].len++;
        }
        cout << "Case " << t << ": " << ClickBox(0, segNum, 0) << endl;
    }
    return 0;
}