中国海洋大学第四届朗讯杯高级组 The Urge to Merge

The Urge to Merge

Time Limit: 1000MS Memory limit: 65536K

题目描述

The Acme Consulting Group has sent you into a new technology park to enhance dynamism, synergy and sustainability. You\'re not sure what any of these terms mean, but you\'re pretty good at making money, which is what you plan on doing. The park consists of a 3 × n grid of facilities. Each facility houses a start-up with an inherent value. By facilitating mergers between neighboring start-ups, you intend to increase their value, thereby allowing you to fulfill your life-long dream of opening your own chain of latte-and-burrito shops.

Due to anti-trust laws, any individual merger may only involve two start-ups and no start-up may be involved in more than one merger. Furthermore, two start-ups may only merge if they are housed in adjacent facilities (diagonal doesn\'t count). The added value generated by a merger is equal to the product of the values of the two start-ups involved. You may opt to not involve a given start-up in any merger, in which case no added value is generated. Your goal is to find a set of mergers with the largest total added value generated. For example, the startup values shown in the figure on the left, could be optimally merged as shown in the figure on the right for a total added value of 171.
                                   

输入

 The first line of each test case will contain a single positive integer n ≤1000 indicating the width of the facilities grid. This is followed by three lines, each containing n positive integers (all ≤ 100) representing the values of each start-up. A line containing a single 0 will terminate input.

输出

 For each test case, output the maximum added value attainable via mergers for that set of start-ups.

示例输入

4
7  2  4  9
3  5  9  3
9  5  1  8
0

示例输出

Case 1: 171

提示

 

来源

 中国海洋大学第四届朗讯杯高级组

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2725

当时比赛的时候看到这道题觉得是道DP，但是因为没有重视高度为3这个条件，想不明白从左上到右下该怎么推，也就没再多考虑。

后来重看这个题，发现整个递推过程只需要从左往右即可。整个dp过程也不难。

以第i列结束的使用的方块，一共有11种形态。不同形态可以从第i-2，第i-1的不同形态转移。

一开始有个地方漏了一点以至于狂WA不止，后来终于发现才AC了。。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int dp[1005][15];
int grid[5][1005];
int n;
int Convers(int v,int a)
{
    switch(v)
    {
    case 1:
        return grid[1][a-1]*grid[1][a];
    case 2:
        return grid[2][a-1]*grid[2][a];
    case 3:
        return grid[3][a-1]*grid[3][a];
    case 4:
        return grid[2][a-1]*grid[2][a]+grid[3][a-1]*grid[3][a];
    case 5:
        return grid[1][a-1]*grid[1][a]+grid[3][a-1]*grid[3][a];
    case 6:
        return grid[1][a-1]*grid[1][a]+grid[2][a-1]*grid[2][a];
    case 7:
        return grid[1][a-1]*grid[1][a]+grid[2][a-1]*grid[2][a]+grid[3][a-1]*grid[3][a];
    case 8:
        return grid[1][a]*grid[2][a];
    case 9:
        return grid[2][a]*grid[3][a];
    case 10:
        return grid[1][a-1]*grid[1][a]+ grid[2][a]*grid[3][a];
    case 11:
        return grid[3][a-1]*grid[3][a]+ grid[1][a]*grid[2][a];
    }
}
int main()
{
    int kase=0;
    while(scanf("%d",&n)==1&&n)
    {
        memset(grid,0,sizeof(grid));
        for(int i=1; i<=3; ++i)
            for(int j=1; j<=n; ++j)
                scanf("%d",&grid[i][j]);
        memset(dp,0,sizeof(dp));
        for(int i=1; i<=n; ++i)
        {
            for(int j=1; j<=11; ++j)
            {
                int res=0;
                if(i>1)
                {
                    switch(j)
                    {
                    case 1:
                        res=max(res,max(dp[i-1][2],max(dp[i-1][3],max(dp[i-1][4],dp[i-1][9]))));
                        break;
                    case 2:
                        for(int k=1; k<=11; ++k)
                            res=max(res,dp[i-2][k]);
                        res=max(res,max(dp[i-1][1],max(dp[i-1][3],dp[i-1][5])));
                        break;
                    case 3:
                        res=max(res,max(dp[i-1][1],max(dp[i-1][2],max(dp[i-1][6],dp[i-1][8]))));
                        break;
                    case 4:
                        res=max(res,dp[i-1][1]);
                        for(int k=1; k<=11; ++k)
                            res=max(res,dp[i-2][k]);
                        break;
                    case 5:
                        res=max(res,dp[i-1][2]);
                        for(int k=1; k<=11; ++k)
                            res=max(res,dp[i-2][k]);
                        break;
                    case 6:
                        res=max(res,dp[i-1][3]);
                        for(int k=1; k<=11; ++k)
                            res=max(res,dp[i-2][k]);
                        break;
                    case 7:
                        for(int k=1; k<=11; ++k)
                            res=max(res,dp[i-2][k]);
                        break;
                    case 8:
                        for(int k=1; k<=11; ++k)
                            res=max(res,dp[i-1][k]);
                        break;
                    case 9:
                        for(int k=1; k<=11; ++k)
                            res=max(res,dp[i-1][k]);
                        break;
                    case 10:
                        res=max(res,max(dp[i-1][2],max(dp[i-1][3],max(dp[i-1][4],dp[i-1][9]))));
                        break;
                    case 11:
                        res=max(res,max(dp[i-1][1],max(dp[i-1][2],max(dp[i-1][6],dp[i-1][8]))));
                        break;
                    }
                }
                if(i==1)
                {
                    if(j==8||j==9)
                        dp[i][j]=Convers(j,i);
                    else
                        dp[i][j]=0;
                }
                else
                    dp[i][j]=res+Convers(j,i);
            }
        }
        int ans=0;
        for(int i=1; i<=11; ++i)
            ans=max(dp[n][i],ans);
        printf("Case %d: ",++kase);
        printf("%d\n",ans);
    }
    return 0;
}