Lightoj 1036(dp)

题意：在一个矿场上有两种矿石，分别对应两个处理他的工厂分别在左边与上边。要求你铺设运输设备每个方格可以向左、上或不运。矿石在运输途中不能转弯，问你最多能运多少矿石到相应的工厂加工。

思路：我们可以注意到由于不能转弯的原因，所以一旦左边的格子被铺上向左的他也也铺上向左的比较有利，否则左边的就白铺了（铺了也运不到）这样每种格子两种选择，就可以写dp方程了：

dp[i][j][0]:当前格子不铺的最大收获

dp[i][j][1]:当前格子向上的最大收获

dp[i][j][2]:当前格子向左的最大收获

F(a, b) : a = max(a, b)

G(i, j): max(dp[i][j][0], dp[i][j][1], dp[i][j][2])

dp[i][j][0]:F(dp[i][j][0], max(G(i-1,j)+mpa[i][j-1], G(i, j-1)+mpb[i-1][j]))

dp[i][j][1]:F(dp[i][j][1], max(G(i, j-1)+mpb[i][j]), dp[i-1][j][1]+mpa[i][j-1]+mpb[i][j]-mpb[i-1][j]))

dp[i][j][2]:F(dp[i][j][2], max(G(i-1, j)+mpa[i][j], dp[i][j-1][2]+mpb[i-1][j]+mpa[i][j]-mpa[i][j-1]))

代码如下：

/**************************************************
 * Author     : xiaohao Z
 * Blog     : http://www.cnblogs.com/shu-xiaohao/
 * Last modified : 2014-05-08 13:35
 * Filename     : L_1035.cpp
 * Description     : 
 * ************************************************/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<unsigned int,unsigned int> puu;
typedef pair<int, double> pid;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;

const int INF = 0x3f3f3f3f;
const double eps = 1E-6;
const int LEN = 1010;
int dp[LEN][LEN][3], n, m, mpa[LEN][LEN], mpb[LEN][LEN];

void F(int &a, int b){
    a = max(a, b);
}

int G(int a, int b){
    int ret = max(dp[a][b][1], dp[a][b][2]);
    ret = max(ret, dp[a][b][0]);
    return ret;
}

int main()
{
//    freopen("in.txt", "r", stdin);

    int T, kase = 1, tmp;
    scanf("%d", &T);
    while(T--){
        scanf("%d%d", &n, &m);
        memset(dp, 0, sizeof dp);
        memset(mpa, 0, sizeof mpa);
        memset(mpb, 0, sizeof mpb);
        for(int i=1; i<=n; i++){
            for(int j=1; j<=m; j++){
                scanf("%d", &tmp);
                mpa[i][j] = mpa[i][j-1] + tmp;
            }
        }
        for(int i=1; i<=n; i++){
            for(int j=1; j<=m; j++){
                scanf("%d", &tmp);
                mpb[i][j] = mpb[i-1][j] + tmp;
            }
        }
        for(int i=1; i<=n; i++){
            for(int j=1; j<=m; j++){
                F(dp[i][j][0], max(G(i-1,j)+mpa[i][j-1], G(i, j-1)+mpb[i-1][j]));
                F(dp[i][j][1], G(i, j-1)+mpb[i][j]);
                F(dp[i][j][2], G(i-1, j)+mpa[i][j]);
                F(dp[i][j][1], dp[i-1][j][1]+mpa[i][j-1]+mpb[i][j]-mpb[i-1][j]);
                F(dp[i][j][2], dp[i][j-1][2]+mpb[i-1][j]+mpa[i][j]-mpa[i][j-1]);
            }
        }
        printf("Case %d: %d
", kase ++, G(n, m));
    }
    return 0;
}