poj3176 Cow Bowling

Cow Bowling

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 13069 Accepted: 8624
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: 

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

5
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
Sample Output

30
Hint

Explanation of the sample: 

The highest score is achievable by traversing the cows as shown above.
Source

USACO 2005 December Bronze

======================================================================================

递归 + 数据记录以减少计算量(Memo-ization, top down)

#include <cstdio>

#define MAXN 360

int InputMat[MAXN][MAXN];	//输入矩阵
int MaxSumAchievable[MAXN][MAXN];
int n;	//输入数据的行数

int myMax(int a, int b)
{
	return (a>b)?a:b;
}

//MaxSumAchievable有可能是0，所以要初始化为负
void init()
{
	for (int i=0; i<MAXN; ++i)
	{
		for (int j=0; j<MAXN; ++j)
		{
			MaxSumAchievable[i][j]=-1;
		}
	}
}

//递归+数据记录
//设f[i][j]为点(i, j)所能有的最大路径和，用一个数组MaxSumAchievable记录已经算出的结果
int f(int i, int j)
{
	if(i==n-1)	
	{
		MaxSumAchievable[i][j]=InputMat[i][j];
		return MaxSumAchievable[i][j];	//最后一行，f值是输入矩阵元素本身，写入记录并返回
	}

	if(MaxSumAchievable[i][j]>=0) return MaxSumAchievable[i][j];	//已经被计算好的记录，返回这个记录

	//若没有记录，用递归函数计算
	MaxSumAchievable[i][j] = InputMat[i][j] + myMax(f(i+1, j), f(i+1, j+1));	//回溯的时候可写入

	return MaxSumAchievable[i][j];
}

int main()
{
	int m=0;

	scanf("%d", &n);
	for (int j=0; j<n; ++j)
	{
		for (int i=0; i<=m; ++i)
		{
			scanf("%d", &InputMat[m][i]);
		}
		++m;
	}

	init();
	printf("%d\n", f(0, 0));

	return 0;
}

======================================================================================

循环，自底向上计算(Tabulation, bottom up)

#include <cstdio>

#define MAXN 360

int a[MAXN][MAXN];	//输入数组

int mymax(int a, int b)
{
	return (a>b)?a:b;
}

int main()
{
	int n;
	scanf("%d", &n);

	for (int i=0; i<n; ++i)
	{
		for (int j=0; j<i+1; ++j)
		{
			scanf("%d", &a[i][j]);
		}
	}

	//bottom-up tabulation
	for (int i=n-2; i>=0; --i)
	{
		for (int j=0; j<i+1; ++j)
		{
			a[i][j] += mymax(a[i+1][j], a[i+1][j+1]);
		}
	}

	printf("%d\n", a[0][0]);

	return 0;
}

======================================================================================

递归打印路径

#include <cstdio>

#define MAXN 360

typedef struct _arr
{
	int info;
	bool isWP;	//means "is way point"
} arr;

arr a[MAXN][MAXN];	//输入数组

int dp[MAXN][MAXN];	//存放答案的数组

//递归打印路径
void PrintRoute(int i, int j, int n)
{
	if(i==n-1)	//递归出口
	{
		printf("%d\n", a[i][j]);
	}
	else
	{
		printf("%d -> ", a[i][j]);
		if(a[i+1][j].isWP) PrintRoute(i+1, j, n);
		else PrintRoute(i+1, j+1, n);
	}
}

int main()
{
	freopen("D:\\in.txt", "r", stdin);
	freopen("D:\\out.txt", "w", stdout);

	int n;
	scanf("%d", &n);

	//读入
	for (int i=0; i<n; ++i)
	{
		for (int j=0; j<i+1; ++j)
		{
			scanf("%d", &(a[i][j].info)); 
			a[i][j].isWP=false;
		}
	}

	//初始化dp数组
	for (int i=0; i<n; ++i)
	{
		dp[n-1][i] = a[n-1][i].info;
	}

	//bottom-up tabulation
	for (int i=n-2; i>=0; --i)
	{
		for (int j=0; j<i+1; ++j)
		{
			if (dp[i+1][j]>dp[i+1][j+1])
			{
				dp[i][j] = a[i][j].info + dp[i+1][j];
				a[i+1][j].isWP=true;
			}
			else
			{
				dp[i][j] = a[i][j].info + dp[i+1][j+1];
				a[i+1][j+1].isWP=true;
			}
		}
	}

	printf("%d\n", dp[0][0]);	//打印结果

	PrintRoute(0, 0, n);	//递归打印路径
	
	return 0;
}

结果：