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; }