POJ1163：The Triangle

The Triangle

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 51019		Accepted: 30914

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



#include
#include
using namespace std;

const int maxn = 1e2 + 2 ;

int a[maxn][maxn] ;
int dp[maxn] ;

int main() {
	ios :: sync_with_stdio(false) ;
	int n ; cin >> n ;
	for(int i = 1 ; i <= n ; i ++) 
		for(int j = 1 ; j <= i ; j ++) 
			cin >> a[i][j] ;
	for(int i = 1 ; i <= n ; i ++) dp[i] = a[n][i] ;
	for(int i = n - 1 ; i >= 0 ; i --) {
		for(int j = 1 ; j <= i ; j ++) {
			dp[j] = max(a[i][j] + dp[j] , a[i][j] + dp[j + 1]) ;
		}
	}
	cout << dp[1] << endl ;
	return 0 ;
}
节省空间做法 ：
#include
#include
using namespace std;
const int maxn = 1e2 + 2 ;
int a[maxn][maxn] ;
int main() {
	ios :: sync_with_stdio(false) ;
	int n ; cin >> n ;
	for(int i = 1 ; i <= n ; i ++) 
		for(int j = 1 ; j <= i ; j ++) 
			cin >> a[i][j] ;
	for(int i = n - 1 ; i >= 0 ; i --) {
		for(int j = 1 ; j <= i ; j ++) {
			a[n][j] = max(a[i][j] + a[n][j] , a[i][j] + a[n][j + 1]) ;
		}
	}
	cout << a[n][1] << endl ;
	return 0 ;
}