POJ 3176 Cow Bowling
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POJ 3176 Cow Bowling
Description
The cows don't use actual bowling balls when they go bowling. They each take a number (in the range 0..99), though, and line up in a standard bowling-pin-like triangle like this:
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
Then the other cows traverse the triangle starting from its tip and moving "down" to one of the two diagonally adjacent cows until the "bottom" row is reached. The cow's score is the sum of the numbers of the cows visited along the way. The cow with the highest score wins that frame.
Given a triangle with N (1 <= N <= 350) rows, determine the highest possible sum achievable.
Input
Line 1: A single integer, N
Lines 2..N+1: Line i+1 contains i space-separated integers that represent row i of the triangle.
Output
Line 1: The largest sum achievable using the traversal rules
Sample Input
573 88 1 02 7 4 44 5 2 6 5
Sample Output
30
Hint
Explanation of the sample:
7
*
3 8
*
8 1 0
*
2 7 4 4
*
4 5 2 6 5
The highest score is achievable by traversing the cows as shown above.
思路分析:
一个典型的动态规划题目,最关键的动态方程为:
result[i][j] = bowling[i][j] + Max(result[i+1][j],result[i+1][j+1]);
代码如下:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#define r 350
#define l 350
using namespace std;
int Max(int a,int b)
{
return a >= b ? a : b;
}
int main()
{
int N;
int bowling[r][l];
int result[r][l];
freopen("1.txt","r",stdin);
scanf("%d",&N);
for(int i = 0;i < N;i++)
{
for(int j = 0;j <= i;j++)
{
scanf("%d",&bowling[i][j]);
}
}
for(int i = 0;i < N;i++)
{
result[N-1][i] = bowling[N-1][i];//最下一层的结果即为原矩阵的最下一层
}
for(int i = N-2;i >= 0;i--)
{
for(int j = 0;j <= i;j++)
{
result[i][j] = bowling[i][j] + Max(result[i+1][j],result[i+1][j+1]);//从下往上不断更新
}
}
printf("%d\n",result[0][0]);
return 0;
}
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