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题目链接： http://www.lightoj.com/volume_showproblem.php?problem=1047
题意：求(p[i][j])上下相邻的 j 不能相同的数塔的最小和。
解法：看代码！
代码：

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <bitset>
#include <math.h>
#include <ctype.h>
#include <time.h>
#include <queue>
#include <map>
#include <set>

using namespace std;

int t;
int n;
int p[25][5];
int dp[25][25];
int pos[25];

int main()
{
    cin >> t;
    for (int ca = 1; ca <= t; ca++)
    {
        memset(dp,0,sizeof(dp));
        cin >> n;
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= 3; j++)
                cin >> p[i][j];

        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= 3; j++)
            {
                if (j == 1)
                    dp[i][j] = min(dp[i - 1][2], dp[i - 1][3]) + p[i][j];
                if (j == 2)
                    dp[i][j] = min(dp[i - 1][1], dp[i - 1][3]) + p[i][j];
                if (j == 3)
                    dp[i][j] = min(dp[i - 1][2], dp[i - 1][1]) + p[i][j];
            }

        cout << "Case " << ca << ": " << min(dp[n][1] ,min(dp[n][2],dp[n][3]))<< endl;
    }
    return 0;
}