UVaLive 3530 Martian Mining (简单DP)

时间：2024-09-18 12:06:02

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;

}

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