2 条题解

  • 1
    @ 2023-8-18 11:29:56
    # include <bits/stdc++.h>
    using namespace std;
    long long n, m;
    long long f[1002][1002][2],a[1002][1002];
    int main() {
    	scanf("%lld%lld", &n, &m);
    	for (long long i = 1; i <= n; i++) {
    		for (long long j = 1; j <= m; j++) {
    			scanf("%lld", &a[i][j]);
    		}
    	}
    	memset(f, 128, sizeof(f));
    	f[1][1][0] = a[1][1];
    	for (long long i = 1; i <= m; i++) {
    		for (long long j = 1; j <= n; j++) {
    			if (i == 1 && j == 1) {
                    continue;
                }
    			f[j][i][0] = max(f[j][i - 1][0], max(f[j][i - 1][1], f[j - 1][i][0])) + a[j][i];
    		}
    		for (long long j = n; j >= 1; j--) {
    			f[j][i][1] = max(f[j][i - 1][0], max(f[j][i - 1][1], f[j + 1][i][1])) + a[j][i];
    		}
    	}
    	printf("%lld", f[n][m][0]);
    	return 0;
    }
    
    • 0
      @ 2022-7-19 21:38:48
      #include <stdio.h>
      typedef long long LL;
      const LL min_ll = -1e18;
      int n, m; LL w[1005][1005], f[1005][1005][2];
      inline LL mx(LL p, LL q, LL r) {return p > q ? (p > r ? p : r) : (q > r ? q : r);}
      inline LL dfs(int x, int y, int from) {
          if (x < 1 || x > n || y < 1 || y > m) return min_ll;
          if (f[x][y][from] != min_ll) return f[x][y][from];
          if (from == 0) f[x][y][from] = mx(dfs(x + 1, y, 0), dfs(x, y - 1, 0), dfs(x, y - 1, 1)) + w[x][y];
          else f[x][y][from] = mx(dfs(x - 1, y, 1), dfs(x, y - 1, 0), dfs(x, y - 1, 1)) + w[x][y];
          return f[x][y][from];
      }
      int main(void) {
      //	freopen("number.in", "r", stdin); freopen("number.out", "w", stdout);
      	scanf("%d %d", &n, &m);
      	for (int i = 1; i <= n; ++i)
      		for (int j = 1; j <= m; ++j) {
      			scanf("%lld", &w[i][j]);
      			f[i][j][0] = f[i][j][1] = min_ll;
      		}
          f[1][1][0] = f[1][1][1] = w[1][1];
      	printf("%lld\n", dfs(n, m, 1));
      	return 0;
      }
      
      
      • 1

      信息

      ID
      124
      时间
      1000ms
      内存
      256MiB
      难度
      2
      标签
      递交数
      25
      已通过
      21
      上传者