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POJ 3176-Cow Bowling(DP||記憶化搜索)

編輯：C++入門知識

POJ 3176-Cow Bowling(DP||記憶化搜索)

Cow Bowling Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 14210 Accepted: 9432

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

Sample Output

數字三角形問題。。可以自底向上坐dp dp[i][j]=ma[i][j]+max(dp[i+1][j],dp[i+1][j+1])

巨水 。。想當初半年前自己懵懵懂懂的刷dp啥都不懂。。哎  真是個悲傷的故事。。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define ll long long
#define maxn 360
#define pp pair
#define INF 0x3f3f3f3f
#define max(x,y) ( ((x) > (y)) ? (x) : (y) )
#define min(x,y) ( ((x) > (y)) ? (y) : (x) )
using namespace std;
int n,dp[maxn][maxn],ma[maxn][maxn];
void solve()
{
	for(int i=0;i=0;i--)
		for(int j=0;j<=i;j++)
			dp[i][j]=max(ma[i][j]+dp[i+1][j],ma[i][j]+dp[i+1][j+1]);
	printf("%d\n",dp[0][0]);
}
int main()
{
	while(~scanf("%d",&n))
	{
		for(int i=0;i
也可以自頂向下記憶化搜索。。然後狀態數組含義都差不多 。。個人覺著搜索比較好寫。。。
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define ll long long
#define maxn 360
#define pp pair
#define INF 0x3f3f3f3f
#define max(x,y) ( ((x) > (y)) ? (x) : (y) )
#define min(x,y) ( ((x) > (y)) ? (y) : (x) )
using namespace std;
int n,dp[maxn][maxn],ma[maxn][maxn];
int dfs(int x,int y)
{
	if(x==n-1)return ma[x][y];
	if(dp[x][y]>=0)return dp[x][y];
	dp[x][y]=0;
	dp[x][y]+=(ma[x][y]+max(dfs(x+1,y),dfs(x+1,y+1)));
	return dp[x][y];
}
int main()
{
	while(~scanf("%d",&n))
	{
		for(int i=0;i