The Triangle

時間限制：1000?ms ?|? 內(nèi)存限制：65535?KB 難度：4 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.

輸入

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.

輸出

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

樣例輸入

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

?

樣例輸出

30

//這個是經(jīng)典的簡單動態(tài)規(guī)劃，典型的多段圖。思路就是建立一個數(shù)組，由下向上動態(tài)規(guī)劃，保存頁子節(jié)點到當前節(jié)點的最大值：? 1 for(int i=num-2;i>=0;i--) //核心 簡單的動規(guī)； 2 { 3 for(int j=0;j<=i;j++) 4 　　{ 5 dp[i][j]+=max(dp[i+1][j],dp[i+1][j+1]); 6 } 7 }

?

1 #include<stdio.h> 2 int main() 3 { 4 int n,i,j,num[110][110]; 5 scanf("%d",&n); 6 for(i=0;i<n;i++) 7 for(j=0;j<=i;j++) 8 scanf("%d",&num[i][j]); 9 for(i=n-2;i>=0;i--) 10 for(j=0;j<=i;j++) 11 { 12 num[i][j]+=num[i+1][j]>num[i+1][j+1]?num[i+1][j]:num[i+1][j+1]; 13 } 14 printf("%d\n",num[0][0]); 15 return 0; 16 }

?

重刷：

#include <cstdio> #include <cstring> #define max(a, b) (a>b?a:b) const int N = 101; int dp[N][N], num[N][N]; int main(){int n;while(scanf("%d", &n) != EOF){memset(dp, 0, sizeof(dp));for(int i = 1; i <= n; i++)for(int j = 1; j <= i; j++)scanf("%d", &num[i][j]);for(int i = 1; i <= n; i++)for(int j = 1; j <= i; j++)dp[i][j]=max(dp[i-1][j], dp[i-1][j-1])+num[i][j];int max = -1;for(int i = 1; i <= n; i++)if(dp[n][i] > max)max = dp[n][i];printf("%d\n", max); }return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/soTired/p/4579513.html

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