joj2511

 2511: Number triangle

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	1s	65536K	427	217	Standard

        5
      3   4
    8   1   2
  5   4   3   6
2   1   7   9   8

A number trangle is composed of N(N<=100) line numbers, the i-th line contains i positive integers(<=100). A chess can walk from the top line of the triangle to the bottom line. Suppose the chess is on the k-th number of one line, then it can only move to the k-th number or the (k+1)-th number of the line below in one step. Find a path from the top to the bottom, which can maximize the sum of the integers on the path.

Input

There are multiple test cases.For each test case, there's an integer N representing the size of the triangle, followed by N lines of positive intergers, the first line has 1 integer, next has two integers... the Nth line has N integers.

Output

The maximum sum of the integers on the path.

Sample Input

3
5
7 0
2 4 3
3
2
6 1
2 3 9

Sample Output

16
12

Developed by skywind, SIYEE

#include<stdio.h>
#include<iostream>
using namespace std;
int a[100+3][100+3];
int s[100+3][100+3];
int main()
{
    int n;
    while(scanf("%d",&n)==1)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=i;j++)
            {
                scanf("%d",&a[i][j]);
            }
        }
        s[1][1]=a[1][1];
        for(int i=2;i<=n;i++)
        {
            s[i][1]=s[i-1][1]+a[i][1];
            for(int j=2;j<=i;j++)
            {
                s[i][j]=max(s[i-1][j],s[i-1][j-1])+a[i][j];
            }
        }
        int maxt=-1;
        for(int i=1;i<=n;i++)
        {
            if(s[n][i]>maxt)
            {
                maxt=s[n][i];
            }
        }
        printf("%d\n",maxt);
    }
    return 0;
}

这个是典型的动态规划的一个题目，关键在于找表达式s[i][j]=max(s[i-1][j-1],s[i-1][j])+a[i][j];表示的是从顶点到第i行第j个点的最大值。。。。。