POJ 1163 The Triangle（三种搜索方式）

DP算法解决最值路径问题

最新推荐文章于 2019-04-17 21:27:20 发布

原创最新推荐文章于 2019-04-17 21:27:20 发布 · 置顶 · 781 阅读

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本文介绍如何使用动态规划（DP）算法解决一个给定三角形数字表中从顶点到基部最大路径和的问题。通过从下向上逐层计算每个节点的最优路径和，最终得出从顶点到基部的最大路径和。

                                                       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

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;
}