poj1163 The Triangle

最新推荐文章于 2020-06-28 17:35:24 发布

转载最新推荐文章于 2020-06-28 17:35:24 发布 · 130 阅读

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原文链接：http://www.cnblogs.com/KingSann/articles/7395747.html

本文介绍了一个经典的动态规划问题——数字三角形最高路径求和。通过递推公式 f[i][j]=max(f[i-1][j],f[i-1][j-1])+a[i][j]，从上到下计算每一步的最大路径和，最终找到从顶部到底部的最大路径总和。

The Triangle

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 50890 Accepted: 30823

Description
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5

(Figure 1)
Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.
Input

Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle, all integers, are between 0 and 99.

Output

Your program is to write to standard output. The highest sum is written as an integer.

Sample Input
5
7
3 8
8 1 0 
2 7 4 4
4 5 2 6 5
Sample Output
30

数字三角形。

f[i][j] = max(f[i - 1][j], f[i - 1][j - 1]) + a[i][j]

 1 #include <cstdio>
 2 #include <algorithm>
 3 
 4 using namespace std;
 5 
 6 int n, f[200][200], ans;
 7 
 8 int main() {
 9   scanf("%d", &n);
10   for(int i = 1 ; i <= n ; i ++) {
11     for(int j = 1 ; j <= i ; j ++) {
12       scanf("%d", &f[i][j]);
13       f[i][j] += max(f[i - 1][j], f[i - 1][j - 1]);
14       if(i == n) {
15         ans = max(ans, f[i][j]);
16       }
17     }
18   }
19   printf("%d\n", ans);
20 }

View Code

如果要强制经过某一点(x, y)？

a[x][y] += INF

答案再减去INF就行

如果禁止进过某一点？

减去INF就行

转载于:https://www.cnblogs.com/KingSann/articles/7395747.html

Time Limit: 1000MS	Memory Limit: 10000K
Total Submissions: 50890	Accepted: 30823