CDOJ 1638 Easy Problem

Problem：http://www.acm.uestc.edu.cn/problem.php?pid=1638

用一個二維數組表示到達某曾某個樓梯口所用的時間，然後從底層向高層DP即可。對每個左邊的樓梯口，上到更高一層後，可以送完所有物品後回到左邊，或者直接去到右邊。

#include <cstdio>

int dp[101][2];
char floo[101];
int end[101][2];

int main()
{
	int T, n, m, top, tmp;
	scanf("%d", &T);
	for(int cases=1; cases<=T; ++cases)
	{
		scanf("%d%d", &n, &m);
		for(int i=1; i<=n; ++i)
		{
			scanf("%s", floo);
			end[i][0] = m-1;
			end[i][1] = 0;
			for(int j=1; j<m-1; ++j)
			{
				if(floo[j] == '*')
				{
					if(j > end[i][1])
						end[i][1] = j;
					if(j < end[i][0])
						end[i][0] = j;
				}
			}
		}
		if(m < 3)
		{
			printf("Case #%d: %d\n", cases, n-1);
			continue;
		}
		for(top = n; top>=1; --top)
		{
			if(end[top][0] == m-1)
				continue;
			else break;
		}
		if(top == 0) 
		{
			printf("Case #%d: %d\n", cases, n-1);
			continue;
		}
		dp[0][0] = dp[0][1] = -1;
		for(int i=1; i<=top; ++i)
		{
			dp[i][0] = dp[i][1] = 0x0fffffff;
			if(end[i][0] != m-1)
			{
				tmp = dp[i-1][0]+1+2*end[i][1];
				if(tmp < dp[i][0])
					dp[i][0] = tmp;
				tmp = dp[i-1][0]+m;
				if(tmp < dp[i][1])
					dp[i][1] = tmp;
				tmp = dp[i-1][1]+1+2*(m-1-end[i][0]);
				if(tmp < dp[i][1])
					dp[i][1] = tmp;
				tmp = dp[i-1][1]+m;
				if(tmp < dp[i][0])
					dp[i][0] = tmp;
			}
			else
			{
				tmp = dp[i-1][0]+1;
				if(tmp < dp[i][0])
					dp[i][0] = tmp;
				tmp = dp[i-1][0]+m;
				if(tmp < dp[i][1])
					dp[i][1] = tmp;
				tmp = dp[i-1][1]+1;
				if(tmp < dp[i][1])
					dp[i][1] = tmp;
				tmp = dp[i-1][1]+m;
				if(tmp < dp[i][0])
					dp[i][0] = tmp;
			}	
		}
		int ans = (dp[top][0]<dp[top][1]?dp[top][0]:dp[top][1]);
		if(top < n) ans += (n-top); 
		printf("Case #%d: %d\n", cases, ans);
	}
	return 0;
}