首页 > 代码库 > projecteuler---->problem=18----Maximum path sum I

projecteuler---->problem=18----Maximum path sum I

2024-07-07 05:25:54   224人阅读

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3
7 4
2 4 6
8 5 9 3

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom of the triangle below:

75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However,Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)


从以下的三角形宝塔的最顶点的数开始，一直往下一层的相邻的数移动，所经过的所有数字的和的最大值是23。

3

7 4

2 4 6

8 5 9 3

亦即3 + 7 + 4 + 9 = 23，暂且称之为最大的「宝塔数」吧。

请找出以下「宝塔」的最大宝塔数：

75

95 64

17 47 82

18 35 87 10

20 04 82 47 65

19 01 23 75 03 34

88 02 77 73 07 63 67

99 65 04 28 06 16 70 92

41 41 26 56 83 40 80 70 33

41 48 72 33 47 32 37 16 94 29

53 71 44 65 25 43 91 52 97 51 14

70 11 33 28 77 73 17 78 39 68 17 57

91 71 52 38 17 14 91 43 58 50 27 29 48

63 66 04 68 89 53 67 30 73 16 69 87 40 31

04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

请注意：由于仅有16384条不同的路径，通过尝试扫描所有的路径我们可以找到答案。不过，Problem 67是一个同样的但要求出一个100层的宝塔的最大宝塔数的问题，它不可能用穷举法解决，因此一个巧妙的算法是必需的。


//============================================================================
// Name        : PE_18.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <cstdio>
using namespace std;

int num[30][30];
int resu=0;
int posX[]={1,1};
int posY[]={0,1};

void getData()
{
    for(int i=0; i < 15; i++){
    	for(int j=0; j < i+1; j++)
    		cin >> num[i][j];
    }
    for(int i=0; i < 15; i++){
    	for(int j=0; j < i+1; j++)
    		cout << num[i][j] << " ";
        cout << endl;
    }

}

void f(int x, int y, int value){
    if(x==15)
    {
    	if(value>resu)
    		resu=value;
    	return;
    }
    f(x+1,y,value+num[x+1][y]);
    f(x+1,y+1, value+num[x+1][y+1]);
}

void solve()
{
    f(0,0,num[0][0]);
    cout << resu << endl;
}

int main() {
//	freopen("18.txt","r",stdin);
    getData();
    solve();
	return 0;
}




联系
我们