POJ1163

The Triangle

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 48800 Accepted: 29478

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

    拿到这个题目，本能的反应——动态规划。思路也很清晰例如以下的这个三角阵
                                   7

                              3         8

                         8         1         0

                   2         7          4         4

              4        5         2          6         5

    边界由于怎么走都只有一种选择，所以不用考虑两条边界。然后继续往下看，我的方法类似于打表，逐层逐个元素计算，懒得打出来了，直接上我的公式b[i][j]=a[i][j]+max(b[i-1][j-1],b[i-1][j])。然后在最后一行找出最大值输出即可（别用排序写=。=）。以下是我的代码

#include <stdio.h>#define M 105int a[M][M],b[M][M];int max(int m,int n){    return m>n?m:n;}int main(){    int n;    //freopen("in.txt","r",stdin);    scanf("%d",&n);    for(int i=0; i<n; i++)        for(int j=0; j<=i; j++)            scanf("%d",&a[i][j]);    b[0][0]=a[0][0];    for(int i=1; i<n; i++)    {        b[i][i]=b[i-1][i-1]+a[i][i];        b[i][0]=b[i-1][0]+a[i][0];    }    for(int i=2; i<n; i++)        for(int j=0; j<i; j++)            b[i][j]=a[i][j]+max(b[i-1][j-1],b[i-1][j]);    int sum=b[n-1][0];    for(int i=1; i<n; i++)        sum=max(b[n-1][i],sum);    printf("%d\n",sum);    return 0;}