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2013
11-12

POJ 3176 Cow Bowling [解题报告] Java

Cow Bowling

问题描述 :

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.

输入:

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.

输出:

Line 1: The largest sum achievable using the traversal rules

样例输入:

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

样例输出:

30

温馨提示:

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.

解题代码:

import java.util.*;   
  
public class Main {   
  
    public static void main(String[] args) {   
        Scanner cin = new Scanner(System.in);   
        int height = cin.nextInt();   
        int[][] tree = new int[height][height];    
        int[][] max = new int[height][height];   
        int maxValue = 0;   
        int left, right = 0;   
           
        for(int i = 0; i < height; i++)   
        {   
            for(int j = 0; j <= i; j++)   
            {   
                tree[i][j] = cin.nextInt();   
//              System.out.print(tree[i][j] + " ");   
            }   
//          System.out.println("\n");   
        }   
           
        max[0][0] = tree[0][0];   
           
  
           
        for(int i = 1; i < height; i++)   
        {   
            for(int j = 0; j <= i; j++)   
            {   
                if(j == 0)   
                    max[i][j] = max[i-1][j] + tree[i][j];   
                else if(j == i)   
                    max[i][j] = max[i-1][j-1] + tree[i][j];   
                else  
                {   
                    if(max[i-1][j-1] >= max[i-1][j])   
                        max[i][j] = max[i-1][j-1] + tree[i][j];   
                    else  
                        max[i][j] = max[i-1][j] + tree[i][j];   
                }   
            }   
        }   
           
        for(int j = 0; j < height; j++)   
        {   
            if(max[height-1][j] > maxValue)   
                maxValue = max[height-1][j];   
        }   
           
        System.out.println(maxValue);   
           
  
    }   
  
}

  1. #include <cstdio>
    #include <algorithm>

    struct LWPair{
    int l,w;
    };

    int main() {
    //freopen("input.txt","r",stdin);
    const int MAXSIZE=5000, MAXVAL=10000;
    LWPair sticks[MAXSIZE];
    int store[MAXSIZE];
    int ncase, nstick, length,width, tmp, time, i,j;
    if(scanf("%d",&ncase)!=1) return -1;
    while(ncase– && scanf("%d",&nstick)==1) {
    for(i=0;i<nstick;++i) scanf("%d%d",&sticks .l,&sticks .w);
    std::sort(sticks,sticks+nstick,[](const LWPair &lhs, const LWPair &rhs) { return lhs.l>rhs.l || lhs.l==rhs.l && lhs.w>rhs.w; });
    for(time=-1,i=0;i<nstick;++i) {
    tmp=sticks .w;
    for(j=time;j>=0 && store >=tmp;–j) ; // search from right to left
    if(j==time) { store[++time]=tmp; }
    else { store[j+1]=tmp; }
    }
    printf("%dn",time+1);
    }
    return 0;
    }