The Triangle
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
题意：给定一个三角形的数字表，求出定点到达最低点的最长的路（点上的值代表其长度）。
解题思路：这是个DP题，从下往上搜，一个点的要么从左边上来，要么从右边上来，所以每次将到达改点的最大值记录下来，状态转换方程为dp[i][j]=max(dp[i+1][j],dp[i+1][j+1])+dp[i][j];
最终输出dp[0][0]即为所求。
复杂度分析： 时间复杂度：o((n)^2)
空间复杂度：o(n^2)
从上往下搜：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
int main()
{
int n;
while(scanf("%d",&n)==1)
{
int dp[n][n];
for(int i=0;i<n;i++)
for(int j=0;j<=i;j++) scanf("%d",&dp[i][j]);
for(int i=1;i<n;i++){
dp[i][0] += dp[i-1][0];
dp[i][i] +=dp[i-1][i-1];
}
for(int i=2;i<n;i++)
for(int j=1;j<=i-1;j++)
dp[i][j]=max(dp[i-1][j],dp[i-1][j-1])+dp[i][j];
for(int i=1;i<n;i++) dp[n-1][0]=max(dp[n-1][0],dp[n-1][i]);
cout << dp[n-1][0] << endl;
}
return 0;
}
或者：从下往上搜
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
int main()
{
int n;
while(scanf("%d",&n)==1)
{
int dp[n][n];
for(int i=0;i<n;i++)
for(int j=0;j<=i;j++) cin>>dp[i][j];
for(int i=n-2;i>=0;i--)
for(int j=0;j<=i;j++)
dp[i][j]=max(dp[i+1][j],dp[i+1][j+1])+dp[i][j];
cout << dp[0][0] << endl;
}
return 0;
}
或者记忆化搜索：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
const int N=105;
int dp[N][N],a[N][N];
int dps(int x,int y){
int L,R;
if(dp[x][y]>=0) return dp[x][y];
L=dps(x+1,y);
R=dps(x+1,y+1);
return dp[x][y]=max(L,R)+a[x][y];
}
int main()
{
int n;
while(scanf("%d",&n)==1)
{
memset(dp,-1,sizeof(dp));
for(int i=0;i<n;i++)
for(int j=0;j<=i;j++)
scanf("%d",&a[i][j]);
for(int i=0;i<n;i++) dp[n-1][i]=a[n-1][i];
dps(0,0);
cout << dp[0][0] << endl;
}
return 0;
}
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