棋盘型动态规划之 CODE[VS] 1220 数字三角形

最新推荐文章于 2025-09-10 15:24:29 发布

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最新推荐文章于 2025-09-10 15:24:29 发布

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CC 4.0 BY-SA版权

文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/shijianming/p/4951732.html

本文介绍了一种使用动态规划方法解决最短路径问题的技术，通过定义状态转移方程和初始化条件，逐步构建最优路径。代码示例中包含了边界检查、状态更新和最终答案获取的过程。

/*
dp[i][j] := 从顶点到达点(i,j)时，所走路径上的最大值
初始化：
	dp[][] = {-INF} // 测试数据有负数，所以初始值 dp[][] = -INF
	dp[1][1] = arr[1][1]
状态方程：
	dp[i][j] = max(dp[i-1][j-1], dp[i-1][j]) + arr[i][j]
答案：
	max{ dp[N][i] } 
	0 <= i <= N-1
*/

 1 #define _CRTDBG_MAP_ALLOC
 2 #include <stdlib.h>
 3 #include <crtdbg.h>
 4 #define _CRT_SECURE_NO_WARNINGS
 5 #define HOME
 6 
 7 #include <iostream>
 8 #include <cstdlib>
 9 #include <cstdio>
10 #include <cstddef>
11 #include <iterator>
12 #include <algorithm>
13 #include <string>
14 #include <locale>
15 #include <cmath>
16 #include <vector>
17 #include <cstring>
18 using namespace std;
19 const int INF = 0x3f3f3f3f;
20 const int MaxN = 30;
21 const int Max = 110;
22 
23 int N;
24 int arr[Max][Max];
25 int dp[Max][Max];
26 
27 bool Check(int x, int y, int len)
28 {
29     if ((x<=0) || (y<=0) || (y>len))
30     {
31         return false;
32     }
33     return true;
34 }
35 
36 void Solve()
37 {
38     int x1, y1, x2, y2;
39     dp[1][1] = arr[1][1];
40     for (int i = 1; i <= N; ++i)
41     {
42         for (int j = 1; j <= i; ++j)
43         {
44             x1 = i - 1;
45             y1 = j - 1;
46             x2 = i - 1;
47             y2 = j;
48             if (Check(x1, y1, i - 1))
49             {
50                 dp[i][j] = max(dp[i][j], dp[x1][y1] + arr[i][j]);
51             }
52             if (Check(x2, y2, i - 1))
53             {
54                 dp[i][j] = max(dp[i][j], dp[x2][y2] + arr[i][j]);
55             }
56         }
57     }
58     /*for (int i = 1; i <= N; ++i)
59     {
60         for (int j = 1; j <= i; ++j)
61         {
62             cout << "(" << i << "," << j << ") : " << dp[i][j] << endl;
63         }
64     }*/
65     cout << *max_element(dp[N], dp[N] + N + 1) << endl;
66 }
67 
68 int main() 
69 {
70 #ifdef HOME
71     freopen("in", "r", stdin);
72     //freopen("out", "w", stdout);
73 #endif
74     for (int i = 0; i < Max; ++i)
75     {
76         for (int j = 0; j < Max; ++j)
77         {
78             dp[i][j] = -INF;
79         }
80     }
81 
82 
83     cin >> N;
84     for (int i = 1; i <= N; ++i)
85     {
86         for (int j = 1; j <= i; ++j)
87         {
88             cin >> arr[i][j];
89         }
90     }
91     Solve();
92 
93 #ifdef HOME
94     cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
95     _CrtDumpMemoryLeaks();
96     system("pause");
97 #endif
98     return 0;
99 }