The Triangle
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 37778 |
|
Accepted: 22685 |
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
Source
IOI 1994
动态规划基础题，主要掌握这种层层递推的思想，由最后一层往上递推；
题意就是从顶层的数字开始，有两个方向开始走，走到底层，输出最大的距离；
#include
#include
#include
using namespace std; const int maxn=101; int dp[maxn][maxn],a[maxn][maxn]; int main() { int n,i,j; scanf("%d",&n); for(i=1;i<=n;i++) for(j=1;j<=i;j++) scanf("%d",&a[i][j]); for(i=1;i<=n;i++) dp[n][i]=a[n][i]; for(i=n-1;i>=0;i--) for(j=1;j<=i;j++) dp[i][j]=max(dp[i+1][j],dp[i+1][j+1])+a[i][j];//往上递推 printf("%d\n",dp[1][1]); return 0; }
还有一种节约内存的写法；
#include
#include
#include
using namespace std; const int maxn=101; int a[maxn][maxn]; int *dp; int main() { int n,i,j; scanf("%d",&n); for(i=1;i<=n;i++) for(j=1;j<=i;j++) scanf("%d",&a[i][j]); dp=a[n]; for(i=n-1;i>=0;i--) for(j=1;j<=i;j++) dp[j]=max(dp[j],dp[j+1])+a[i][j]; printf("%d\n",dp[1]); return 0; }