POJ 1163 The Triangle

The Triangle

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 47093 Accepted: 28499

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

Source

IOI 1994

一道简单的动态规划题目，状态转移方程为：dp[i][j] = max(dp[i - 1][j], dp[i-1][j-1])+a[i][j];

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;int a[105][105],dp[105][105];int main(){int n;while(~scanf("%d",&n)){memset(a,0,sizeof(a));memset(dp,0,sizeof(dp));for(int i = 1; i <= n; i ++){for(int j = 1; j <= i; j ++)scanf("%d",&a[i][j]);}for(int i = 1; i <= n; i ++){for(int j = 1; j <= i; j ++){dp[i][j] = max(dp[i - 1][j],dp[i - 1][j - 1]) + a[i][j];}}int Max = 0;for(int i = 1; i <= n; i ++)Max = max(Max,dp[n][i]);printf("%d\n",Max);}return 0;}