The Triangle--poj--1163

The Triangle

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 19122		Accepted: 11144

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

Source

IOI 1994

解题思路：

基本的DP，从底往上扫描数组，递推式如下：

a（i，j）=a（i，j）+max（a（i+1，j），a（i+1，j+1））

其实刚开始用的是dfs递归，但结果操时。

先贴上dfs的代码吧(TLE)：

#include <iostream>

const int MAX = 150;

int N;

int in[MAX][MAX];

//深搜

int dfs(int i, int j)

{

if(i == N-1) //到底了

{

return in[i][j];

}

int tm1 = dfs(i+1, j) + in[i][j]; //当前节点的左下边的节点

int tm2 = dfs(i+1, j+1) + in[i][j]; //当前节点的右下边节点

return tm1>tm2 ? tm1 : tm2; //取大的咯

}

int main()

{

int max = 0;

std::cin>>N;

for(int i=0; i<N; i++)

{

for(int j=0; j<i+1; j++)

{

std::cin>>in[i][j];

}

}

std::cout<<dfs(0, 0)<<std::endl; //从根节点开始

system("pause");

return 0;

}

下面是DP的代码（AC）：

#include <iostream>

const int MAX = 101;

int N;

int in[MAX][MAX];

int main()

{

int max = 0;

std::cin>>N;

for(int i=0; i<N; i++)

{

for(int j=0; j<i+1; j++)

{

std::cin>>in[i][j];

}

}

//从数组倒数第二行开始扫描,从底向上计算

//in[i,j] = in[i,j] + max(in[i+1, j], in[i+1, j+1])

//最后最大值就存于in[0][0]中

for(int i=N-2; i>=0; i--)

{

for(int j=0; j<i+1; j++)

{

if(in[i+1][j] > in[i+1][j+1])

in[i][j] += in[i+1][j];

else

in[i][j] += in[i+1][j+1];

}

}

std::cout<<in[0][0]<<std::endl;

system("pause");

return 0;

}