1163:The Triangle（动态规划）

 

• 查看
• 提交
• 统计
• 提示
• 提问

总时间限制: 

1000ms

 

内存限制: 

65536kB

描述

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.

输入

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.

输出

Your program is to write to standard output. The highest sum is written as an integer.

样例输入

5
7
3 8
8 1 0 
2 7 4 4
4 5 2 6 5

样例输出

30

来源

IOI 1994

当N很大时，递归会超时

递归：

#include <bits/stdc++.h>
using namespace std;
int a[105][105],n;
int maxsum(int i,int j) {
    if(i==n) {
        return a[i][j];
    }
    int sum1=maxsum(i+1,j);
    int sum2=maxsum(i+1,j+1);
    return a[i][j]+max(sum1,sum2);
}
int main() {
    cin>>n;
    for(int i=1; i<=n; i++) {
        for(int j=1; j<=i; j++) {
            cin>>a[i][j];
        }
    }
    cout<<maxsum(1,1);

    return 0;
}

所以选择动态规划，在每一步求得最大值时记录下来，下次就无须再次计算。

#include <bits/stdc++.h>
using namespace std;
int a[105][105],n;
int ms[105][105];
int maxsum(int i,int j) {
    if(i==n) {
        return a[i][j];
    }
    if(ms[i+1][j]==-1) {
        ms[i+1][j]=maxsum(i+1,j);
    }
    if(ms[i+1][j+1]==-1) {
        ms[i+1][j+1]=maxsum(i+1,j+1);
    }
    return a[i][j]+max(ms[i+1][j],ms[i+1][j+1]);
}
int main() {
    cin>>n;
    memset(ms,-1,sizeof(ms));
    for(int i=1; i<=n; i++) {
        for(int j=1; j<=i; j++) {
            cin>>a[i][j];
        }
    }
    cout<<maxsum(1,1);

    return 0;
}