数值三角形的一个递归解法(超时)

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The Triangle

Time Limit: 1000ms, Special Time Limit:2000ms, Memory Limit:32768KB

Total submit users: 162, Accepted users: 153

Problem 10014 : No special judgement

Problem 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 T, the number of test cases, for each test case: the first line contain a 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 for each test case one line.

Sample Input

Sample Output

Problem Source

IOI 1994

自下而上相加:

#include <iostream>
using namespace std;
#define MAX 110
#define X -1

int visitnode(int num[][MAX],int i ,int j);
main()
{
int i=0,j=0;
int n=0,k=0,total=0;
int num[MAX][MAX];
scanf("%d",&total); //number of cases

for(k=0;k<total;k++){
     scanf("%d",&n);
     for(i=0;i<n;i++)
         for(j=0;j<=i;j++)scanf("%d",&num[i][j]);    //input sum
     for(i=0;i<=n;i++)num[n][i]=X;
     /*for(i=0;i<=n;i++){
         for(j=0;j<=i;j++)printf("%d ",num[i][j]);
         printf(" ");
     } */
     for(i=n-1;i>0;i--){
          for(j=0;j<n;j++){
             num[i-1][j]+=max(num[i][j],num[i][j+1]);
          }
     }
     /*
     for(i=0;i<=n;i++){
         for(j=0;j<=i;j++)printf("%d ",num[i][j]);
         printf(" ");
     }
     */
     printf("%d ",num[0][0]);
}

return 0;
}

递归解法(超时):

#include <stdio.h>
#define MAX 110
#define X -1

int visitnode(int num[][MAX],int i ,int j);
main()
{
int i=0,j=0;
int n=0,k=0,total=0;
int num[MAX][MAX];
scanf("%d",&total); //number of cases

for(k=0;k<total;k++){
     scanf("%d",&n);
     for(i=0;i<n;i++)
         for(j=0;j<=i;j++)scanf("%d",&num[i][j]);    //input sum
     for(i=0;i<=n;i++)num[n][i]=X;
     /*for(i=0;i<=n;i++){
         for(j=0;j<=i;j++)printf("%d ",num[i][j]);
         printf(" ");
     }    */
     printf("%d ",visitnode(num,0,0));
}

system("PAUSE");
}

int visitnode(int num[][MAX],int i ,int j)
{
    int a=0,b=0;
    if(num[i][j]==X)return 0;
    a=num[i][j]+visitnode(num,i+1,j);
    b=num[i][j]+visitnode(num,i+1,j+1);
    return a>b?a:b;
}