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题目链接：hdu 4826 Labyrinth

题目大意：中文题。

解题思路：不难想的递推，dp[i][j][0]从上面过来的情况，dp[i][j][1]从下面过来的情况，然后这两种情况都可以从前一列走过来。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int N = 105;
const int INF = 0x3f3f3f3f;
int n, m, g[N][N], dp[N][N][2];

void init () {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            scanf("%d", &g[i][j]);
}

int solve () {

    for (int i = 1; i <= m; i++)
        dp[0][i][0] = dp[n+1][i][1] = -INF;

    dp[0][1][0] = 0;
    for (int i = 1; i <= n; i++) {
        dp[i][1][0] = dp[i-1][1][0] + g[i][1];
        dp[i][1][1] = -INF;
    }

    for (int j = 2; j <= m; j++) {

        for (int i = 1; i <= n; i++)
            dp[i][j][0] = max(dp[i-1][j][0], max(dp[i][j-1][0], dp[i][j-1][1])) + g[i][j];

        for (int i = n; i >= 1; i--)
            dp[i][j][1] = max(dp[i+1][j][1], max(dp[i][j-1][0], dp[i][j-1][1])) + g[i][j];
    }
    return max(dp[1][m][0], dp[1][m][1]);
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int i = 1; i <= cas; i++) {
        init ();
        printf("Case #%d:\n%d\n", i, solve());
    }
    return 0;
}

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hdu 4826 Labyrinth(dp)

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原文地址：http://blog.csdn.net/keshuai19940722/article/details/26517843