UVaLive 3530 Martian Mining (简单DP)

题意：给定一个n*m的网格，每个格子里有A矿和B矿数量，A必须由右向左运，B只能从下向上运，中间不能间断，问最大总数量。

析：一个简单DP，dp[i][j] 表示 从 （0， 0） 到 （i， j） 最大人运输量。要么向左运输，要么向上运输，取最大值。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 500 + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int dp[maxn][maxn];
int a[maxn][maxn];
int b[maxn][maxn];

int main(){
    while(scanf("%d %d", &n, &m) == 2 && n + m){
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= m; ++j){
                scanf("%d", &a[i][j]);
                a[i][j] += a[i][j-1];
            }
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= m; ++j){
                scanf("%d", &b[i][j]);
                b[i][j] += b[i-1][j];
            }

        memset(dp, 0, sizeof dp);
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= m; ++j)
                dp[i][j] = max(dp[i-1][j] + a[i][j], dp[i][j-1] + b[i][j]);
        printf("%d\n", dp[n][m]);
    }
    return 0;
}