POJ 1163 数塔 — 基础DP

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原文链接：http://www.cnblogs.com/mcgrady_ww/p/6657083.html

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#java

本文介绍了一个经典的编程问题——寻找三角形网格中从顶点到底部的最大路径和，并提供了两种不同的C++实现方案。该问题通常出现在编程竞赛中，通过动态规划的方法高效求解。

The Triangle

Time Limit: 1000 MS Memory Limit: 10000 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

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

Sample Output

//poj 1163
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn = 105;

int a[maxn][maxn];
int dp[maxn][maxn];

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(a,0,sizeof(a));
        memset(dp,0,sizeof(dp));

        for(int i = 1 ; i <= n; i++)
            for(int j = 1; j <= i; j++)
                scanf("%d",&a[i][j]);

        for(int i = n; i > 0; i--)
            for(int j = 1; j <= i; j++)
                dp[i][j] = a[i][j] + max(dp[i+1][j],dp[i+1][j+1]);
        printf("%d\n",dp[1][1]);
    }
    return 0;
}

优化代码

#include<iostream>
#include<stdio.h>
using namespace std;
#define maxn 105

int a[maxn][maxn];
int b[maxn];

int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= i; j++)
            scanf("%d",&a[i][j]);

    for(int i = 1; i <= n; i++)
            b[i] = a[n][i]; //只在最后一行上不断更新；
//    for(int i = 1; i <= n; i++)
//        printf("--%d--\n",b[i]);
    for(int i = n - 1; i > 0; i--)
        for(int j = 1; j <= i; j++)
            b[j] = max(b[j],b[j+1])+a[i][j];
    printf("%d\n",b[1]);

    return 0;
}