POJ 1163 The Triangle

最新推荐文章于 2019-01-09 16:39:02 发布

ssSky_x

最新推荐文章于 2019-01-09 16:39:02 发布

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分类专栏： ACM之路--动态规划篇文章标签： ACM DP POJ1163

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The Triangle

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 48243		Accepted: 29173

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

Source

IOI 1994

---------------------------------华丽的分割线-------------------------------

解题思路：
题目是求这个数字三角形从上到下（或者相反）的一条路径的最大和。
状态转移方程：
DP[I][J]=MAX(DP[I+1][J],DP[I+1][J+1])+POINT[I][J].

Code:

#include<stdio.h>
#include<cstring>
int tri[102][102];
int dp[102][102];
int i,j,k,n,m,p,t;
int main()
{
    memset(tri,0,sizeof(tri));
    memset(dp,0,sizeof(dp));

    scanf("%d",&n);

    for(i=1;i<=n;i++)
        for(j=1;j<=i;j++)
            scanf("%d",&(tri[i][j]));
    for(i=n;i>=1;i--)
        for(j=n-(n-i);j>=1;j--)
            dp[i][j]=dp[i+1][j]>dp[i+1][j+1]?dp[i+1][j]+tri[i][j]:dp[i+1][j+1]+tri[i][j];
    printf("%d\n",dp[1][1]);
    return 0;
}