(记忆化搜索 )The Triangle--hdu --1163

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

Sample Output

30


单纯的递归， 但是会超时

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

#define N 110
#define max(a,b) (a>b?a:b)

int a[N][N];

int DFS(int x, int y, int n)
{
    if(x>n || y>n)
        return 0;

    if(x==n && y==n)
        return a[x][y];

    return a[x][y]+ max(DFS(x+1,y, n), DFS(x+1, y+1, n));
}

int main()
{
    int n;

    while(scanf("%d", &n)!=EOF)
    {
        int i, j;

        memset(a, 0, sizeof(a));

        for(i=1; i<=n; i++)
        for(j=1; j<=i; j++)
            scanf("%d", &a[i][j]);

        printf("%d
", DFS(1,1,n));
    }
    return 0;
}

用上记忆化搜索后，不会超时了

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

#define N 110
#define max(a,b) (a>b?a:b)

int a[N][N], dp[N][N];

int  DFS(int x, int y, int n)
{
    if(x>n || y>n)
        return 0;

    if(dp[x][y]!=-1)
        return dp[x][y];
    else
    {
        if(x==n && y==n)
        return a[x][y];

        dp[x+1][y] = DFS(x+1, y, n);
        dp[x+1][y+1] = DFS(x+1, y+1, n);

        return a[x][y]+ max(dp[x+1][y], dp[x+1][y+1]);
    }
}

int main()
{
    int n;

    while(scanf("%d", &n)!=EOF)
    {
        int i, j;

        memset(a, 0, sizeof(a));
        memset(dp, -1, sizeof(dp));

        for(i=1; i<=n; i++)
        for(j=1; j<=i; j++)
            scanf("%d", &a[i][j]);

        printf("%d
", DFS(1, 1, n));
    }
    return 0;
}

勿忘初心