poj1163 the triangle 题解

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Description

73   88   1   02   7   4   44   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

573 88 1 0 2 7 4 44 5 2 6 5

Sample Output

题目大意：求出三角形的最大和，规定路径从第一行第一个为起点，路径为向下一行的该列或者下一行的右边的一列（在所有数字向左对齐的情况下），终点为最后一行的某个数。

解题思路：这是一道dp教程中十分常见的入门题，在解这道题目我们要尽量防止回溯，因为很有可能会超时（实际上也超了），所以我们十分需要用一个数组来储存过程中计算得到的结果，在代码里面我定义为dp数组，dp[i][j]表示从倒数第一排开始走到i排j列的最大和，如果输入是样例输入的话，那么dp数组应该是这样的。

3023 2120 13 107 12 10 104 5 2 6 5

（其余为0）

递归代码

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#define MAXN 100using namespace std;int map[MAXN + 5][MAXN + 5];int dp[MAXN + 5][MAXN + 6];int n;int sum;int maxsum(int x, int y){    if(dp[x][y] != -1)//dp[x][y]!=-1表示已经计算了，不必再算        return dp[x][y];    if(x == n)//x==n表示最后一行，最小和为map[x][y]他本身    {        dp[x][y] = map[x][y];    }    else    {        int i = maxsum(x + 1, y);        int j = maxsum(x + 1, y + 1);        dp[x][y] = max(i, j) + map[x][y];    }    return dp[x][y];}int main(){    cin >> n;    for(int i = 1; i <= n; i ++)    {        for(int j = 1; j <= i; j ++)        {            scanf("%d", &map[i][j]);            dp[i][j] = -1;        }    }    cout << maxsum(1, 1) << endl;    for(int i = 1; i <= n; i++)    {        for(int j = 1; j <= i; j ++)        {            cout << dp[i][j] << ' ';        }        cout << endl;    }    return 0;}/*573 88 1 02 7 4 44 5 2 6 5ans:30*/

递推代码

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#define MAXN 100using namespace std;int map[MAXN + 5][MAXN + 5];int dp[MAXN + 5][MAXN + 6];int n;int sum;int main(){    cin >> n;    for(int i = 1; i <= n; i ++)    {        for(int j = 1; j <= i; j ++)        {            scanf("%d", &map[i][j]);        }    }    for(int i = 1; i <= n; i ++)//最后一行的最大和是最后一行的数字本身        dp[n][i] = map[n][i];    for(int i = n - 1; i >= 1; i --)//从最后一排到第一排    {        for(int j = 1; j <= i; j ++)        {            dp[i][j] = max(dp[i + 1][j], dp[i + 1][j + 1]) + map[i][j];        }    }    cout << dp[1][1] << endl;    return 0;}/*573 88 1 02 7 4 44 5 2 6 5ans:30*/
递推代码的空间优化：

空间优化的思路：将dp从二维转换为一维很明显能节省很多空间，dp[i]这时候表示第一行i列为终点的最大和

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#define MAXN 100using namespace std;int map[MAXN + 5][MAXN + 5];int dp[MAXN + 5];int n;int sum;int main(){    cin >> n;    for(int i = 1; i <= n; i ++)    {        for(int j = 1; j <= i; j ++)        {            scanf("%d", &map[i][j]);        }    }    for(int i = 1; i <= n; i ++)//最后一行的最大和是最后一行的数字本身    {        dp[i] = map[n][i];    }    for(int i = n - 1; i >= 1; i --)//从最后一排到第一排    {        for(int j = 1; j <= i; j ++)        {            dp[j] = max(dp[j], dp[j + 1]) + map[i][j];        }    }    cout << dp[1] << endl;    return 0;}/*573 88 1 02 7 4 44 5 2 6 5ans:30*/

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