POJ 1163 The Triangle

链接：http://poj.org/problem?id=1163

The Triangle

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 41060 Accepted: 24800

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

Source
IOI 1994

大意——给定一个n层的三角形数字塔，从顶到底只能从两个方向向下走，从左或从右，问：从顶到底路径数值和最大的那个数。

思路——显然这是一个最简单的dp问题。我们从顶向下递推即可。边界点(i,0)只有一条路从(i-1,0)过来，则dp[i][0] = dp[i-1][0] + a[i][0]；边界点(i,i)也只有一条路从(i-1,i-1)过来，则dp[i][i] = dp[i-1][i-1] + a[i][i]；对于非边界点(i,j),它只能从(i-1,j-1)或者(i-1,j)过来,则dp[i][j]=max{dp[i-1][j-1],dp[i-1][j]} +a[i][j]。因此，结果即为max{dp[n-1][j]|j=0,1,...,n-1}。

复杂度分析——时间复杂度：O(n^2),空间复杂度：O(n^2)

附上AC代码：

#include <iostream>#include <cstdio>#include <string>#include <cmath>#include <iomanip>#include <ctime>#include <climits>#include <cstdlib>#include <cstring>#include <algorithm>using namespace std;typedef unsigned int UI;typedef long long LL;typedef unsigned long long ULL;typedef long double LD;const double PI = 3.14159265;const double E = 2.71828182846;const short MAX = 100;int max(int a, int b);int main(){ios::sync_with_stdio(false);short num[MAX][MAX];int dp[MAX][MAX];int N;while (cin >> N){for (int i=0; i<N; i++)for (int j=0; j<=i; j++)cin >> num[i][j];for (int i=0; i<N; i++)for (int j=0; j<=i; j++){if (0 == i)dp[i][j] = num[i][j];else if (0 == j)dp[i][j] = dp[i-1][j]+num[i][j];else if (i == j)dp[i][j] = dp[i-1][j-1]+num[i][j];elsedp[i][j] = max(dp[i-1][j], dp[i-1][j-1])+num[i][j];}int ans = dp[N-1][0];for (int i=1; i<N; i++)ans = max(ans, dp[N-1][i]);cout << ans << endl;}return 0;}int max(int a, int b){if (a > b)return a;return b;}