洛谷 P1006 [NOIP2008 提高组] 传纸条

合集 - 3.26~4.13 蓝桥杯省赛备赛(28)

1.P8687 [蓝桥杯 2019 省 A] 糖果2024-03-252.洛谷 P1115 最大子段和2024-03-253.写模板， 并查集。2024-03-254.洛谷 P1656 炸铁路2024-03-265.写模板，树状数组。2024-03-266.洛谷 P3374 【模板】树状数组 12024-03-267.洛谷 P3368 【模板】树状数组 22024-03-268.洛谷 P9237 [蓝桥杯 2023 省 A] 像素放置2024-03-269.洛谷 P5937 [CEOI1999] Parity Game2024-03-2710.洛谷 P8306 【模板】字典树2024-03-2711.写模板，kmp。2024-03-2712.写模板， LCA。2024-03-2813.洛谷 P1967 [NOIP2013 提高组] 货车运输2024-03-2814.写模板，线段树2024-03-2815.蓝桥杯-百亿富翁2024-03-2916.洛谷 P3372 【模板】线段树 12024-03-2917.模拟比赛-14届研究生组C++省赛2024-04-0118.写模板， 线性筛2024-04-0119.写模板-质数的简单应用2024-04-0220.洛谷 P1196 [NOI2002] 银河英雄传说2024-04-0421.洛谷 P1004 [NOIP2000 提高组] 方格取数2024-04-0422.蓝桥杯 计算方程2024-04-04

23.洛谷 P1006 [NOIP2008 提高组] 传纸条2024-04-05

24.蓝桥杯-算法赛第9场强者：贝贝的2.02024-04-0825.P1002 [NOIP2002 普及组] 过河卒2024-04-0826.蓝桥杯，推导部分和2024-04-0827.牛客，小红不想做完全背包2024-04-0928.蓝桥杯-岛屿个数2024-04-10

收起

题意：传纸条，跟方格取数一样，但是两条路径不能有重复的。

思路：还是一样的走，但是x1跟x2不能相等，包括现在跟上一个状态。

总结：看了题解，发现题解大多数都是逻辑不正确的，更有离谱的是数组范围都不加特判，数组访问越界但是可以ac的情况，数据太烂了，放个自以为正确的思路吧，发现之前自己提交的满分代码也不是逻辑完全正确，放个自以为正确的代码。

void solve(){
	int n, m;
	cin >> n >> m;

	vector<vector<int>> grid(n + 1, vector<int> (m + 1, 0));
	for (int i = 1; i <= n; ++i){
		for (int j = 1; j <= m; ++j){
			cin >> grid[i][j];
		}
	}

	vector<vector<vector<int>>> dp(2, vector<vector<int>> (n + 1, vector<int>(m + 1, 0)));
	dp[0][1][1] = grid[1][1];

	vector<vector<int>> zeros(n + 1, vector<int>(m + 1, 0));
	int cur = 0;
	for (int s = 1; s <= n + m - 2; ++s){
		cur ^= 1;
		for (int x1 = 1; x1 <= min(s + 1, n); ++x1){
			for (int x2 = 1; x2 <= min(s + 1, n); ++x2){
				if (x1 == x2 && (s != n + m - 2)){
					continue;
				}
				int y1 = s + 2 - x1;
				int y2 = s + 2 - x2;
				if (max(y1, y2) > m){
					continue;
				}
				if (x1 > 1 && x2 > 1){
					dp[cur][x1][x2] = max(dp[cur][x1][x2], dp[!cur][x1 - 1][x2 - 1] + grid[x1][y1] + grid[x2][y2]);
				}
				if (x1 > 1 && y2 > 1 && (x1 - 1 != x2 || (x1 == 2 && s == 1) || s == n + m - 2)){
					dp[cur][x1][x2] = max(dp[cur][x1][x2], dp[!cur][x1 - 1][x2] + grid[x1][y1] + grid[x2][y2]);
				}
				if (y1 > 1 && x2 > 1 && (x2 - 1 != x1 || (x2 == 2 && s == 1) || s == n + m - 2)){
					dp[cur][x1][x2] = max(dp[cur][x1][x2], dp[!cur][x1][x2 - 1] + grid[x1][y1] + grid[x2][y2]);
				}
				if (y1 > 1 && y2 > 1){
					dp[cur][x1][x2] = max(dp[cur][x1][x2], dp[!cur][x1][x2] + grid[x1][y1] + grid[x2][y2]);
				}
			}
		}
	}
  //  for (int i = 1; i <= n; ++i){
	  //  for (int j = 1; j <= m; ++j){
	  //      cout <<dp[cur][i][j] << " \n"[j == m];
	   // }
	//}

	cout << dp[cur][n][m]  << '\n';
}

真是个sb，又调了一下，ac了。思路不变，但是dp的状态搞错了，应该输出dp[cur][n][n],哎，服了！

void solve(){
    int n, m;
    cin >> n >> m;

    vector<vector<int>> grid(n + 1, vector<int> (m + 1, 0));
    for (int i = 1; i <= n; ++i){
        for (int j = 1; j <= m; ++j){
            cin >> grid[i][j];
        }
    }

    vector<vector<vector<int>>> dp(2, vector<vector<int>> (n + 1, vector<int>(n + 1)));
    dp[0][1][1] = grid[1][1];

    int cur = 0;
    for (int s = 1; s <= n + m - 2; ++s){
        cur ^= 1;
        for (int x1 = 1; x1 <= min(s + 1, n); ++x1){
            for (int x2 = 1; x2 <= min(s + 1, n); ++x2){
                if (x1 == x2 && (s != n + m - 2)){
                    continue;
                }
                int y1 = s + 2 - x1;
                int y2 = s + 2 - x2;
                if (max(y1, y2) > m){
                    continue;
                }
                if (x1 > 1 && x2 > 1){
                    dp[cur][x1][x2] = max(dp[cur][x1][x2], dp[!cur][x1 - 1][x2 - 1] + grid[x1][y1] + grid[x2][y2]);
                }
                if (x1 > 1 && y2 > 1 && (x1 - 1 != x2 || (s == 1) || s == n + m - 2)){
                    dp[cur][x1][x2] = max(dp[cur][x1][x2], dp[!cur][x1 - 1][x2] + grid[x1][y1] + grid[x2][y2]);
                }
                if (y1 > 1 && x2 > 1 && (x2 - 1 != x1 || (s == 1) || s == n + m - 2)){
                    dp[cur][x1][x2] = max(dp[cur][x1][x2], dp[!cur][x1][x2 - 1] + grid[x1][y1] + grid[x2][y2]);
                }
                if (y1 > 1 && y2 > 1){
                    dp[cur][x1][x2] = max(dp[cur][x1][x2], dp[!cur][x1][x2] + grid[x1][y1] + grid[x2][y2]);
                }
            }
        }
    }
  //  for (int i = 1; i <= n; ++i){
      //  for (int j = 1; j <= m; ++j){
      //      cout <<dp[cur][i][j] << " \n"[j == m];
       // }
    //}
    cout << dp[cur][n][n]  << '\n';
}

而且dp的大小也定义错了。。明明应该是按行数来定义，结果把列也加进去了，直觉错误了！！