POJ——Problem1163（三角形动态规划）

The Triangle

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 52111 Accepted: 31468

Description

73   88   1   02   7   4   44   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

573 88 1 0 2 7 4 44 5 2 6 5

Sample Output

30

解题思路

简单的翻译一下，这道题类似于buaaOJ中的掉坑系列问题，就是从三角形的顶端开始，每一步只能是向下或者右下，求走到底边时所能获得的最大金币之和。

这里用d[i][j]表示点（i，j）所放的金币数，用max[i][j]表示从点（i，j）走到底边所能获得的最大金币数。

所以可以求得状态转移方程为：

maxSum[i][j]=①d[i][j],i=n

          ②max(maxSum[i+1][j],maxSum[i+1][j+1])+d[i][j]

示例代码

#include<cstdio>#include<cstring>int max(int a,int b){    if(a>b)        return a;    return b;}int d[101][101];int n;int maxSum[101][101];int main(){    scanf("%d",&n);    for(int i=1;i<=n;i++)        for(int j=1;j<=i;j++)            scanf("%d",&d[i][j]);    for(int i=1;i<=n;i++)        maxSum[n][i]=d[n][i];    for(int i=n-1;i>=1;i--)        for(int j=1;j<=i;j++)            maxSum[i][j]=max(maxSum[i+1][j],maxSum[i+1][j+1])+d[i][j];    printf("%d",maxSum[1][1]);}