2017年上海金馬五校程序設計競賽：Problem K : Treasure Map

Problem K : Treasure Map

From: DHUOJ, 2017060311

 (Out of Contest)

Time Limit: 3 s

---

Description

There is a robot, its task is to bury treasures in order on a N × M grids map, and each treasure can be represented by its weight, which is an integer.

The robot begins to bury the treasures from the top-left grid. Because it is stupid, it can only Go straight until the border or the next grid has already been buried a treasure, and then it turns right.

Its task is finished when all treasures are buried. Please output the treasure map as a N × M matrix.

 

Input

There are several test cases, each one contains two lines.
First line: two integers N and M (1 ≤ N, M ≤ 100), indicate the size of the map.
Second line: N × M integers, indicate the weight of the treasures in order.

 

Output

For each test case, output a N × M matrix contains the weight of the treasures buried by the robot. There is one space between two integers.

 

Sample Input

2 2
3 2 1 4
3 3
1 2 3 4 5 6 7 8 9

 

Sample Output

3 2
4 1
1 2 3
8 9 4
7 6 5

題目意思就是給定一個n*m的數組，然後按照蛇形填數的形式填進去，最後輸出即可。。

然後我就按照最簡單的蛇形填數來做了，測試數據都過了呀，親，然後就是超時超時超時~~~~~TLE TLE TLE  0.0

附上超市的代碼：

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int a[111][111];
int main()
{
    int n,m,count,x,y,b[10001];
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=1; i<=n*m; i++)
        {
            scanf("%d",&b[i]);
        }
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
            a[i][j]=0;
        a[x=0][y=0]=b[1];
        count =  1;
        while(count<n*m)
        {
            while(y+1<m &&!a[x][y+1])  a[x][++y] =b[++count];
            while(x-1>=0 &&!a[x-1][y]) a[--x][y]=b[++count];
            while(y-1>=0 &&!a[x][y-1]) a[x][--y]=b[++count];
            while(x+1<n && !a[x+1][y]) a[++x][y]=b[++count];
        }
        for(x=0; x<n; x++)
        {
            for(y=0; y<m; y++)
            {
                if(y==m-1)
                    printf("%d",a[x][y]);
                else
                printf("%d ",a[x][y]);
            }
            printf("\n");
        }
    }
}

不知道爲啥超時呀，感覺基本原理都是這樣的。。。。。

然後dalao   AC的代碼在此：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;

const int maxn = 102;
int N,M;
int Map[maxn][maxn];
int a[maxn*maxn];
int main()
{
    int up,down,left,right;     ///分別記錄上邊界，下邊界，左邊界，右邊界
    while(~scanf("%d%d",&N,&M))
    {
        for(int i = 1; i <= N*M; i++)
            scanf("%d",&a[i]);
        up = 0;
        down = N+1;
        left = 0;
        right = M+1;
        int ccount = 1;
        while(1)
        {
            ///先向右填數
            if(up+1<down) ///必須要判斷，上下邊界相鄰，不能填數了。同理左右邊界相鄰也不能填數了。
            {
                for(int i = left+1; i<right; i++)
                {
                    Map[up+1][i] = a[ccount++];
                }
                up++;
            }
            else break;
            ///然後往下走
            if(left<right-1)
            {
                for(int i = up+1; i < down; i++)
                {
                    Map[i][right-1] = a[ccount++];
                }
                right--;
            }
            else break;
            ///然後往左走
            if(up<down-1)
            {
                for(int i = right-1; i > left; i--)
                {
                    Map[down-1][i] = a[ccount++];
                }
                down--;
            }
            else break;
            ///然後往上走
            if(left+1<right)
            {
                for(int i = down-1; i > up; i--)
                {
                    Map[i][left+1] = a[ccount++];
                }
                left++;
            }
            else break;
        }
        for(int i = 1; i <= N; i++)
        {
            for(int j = 1; j <= M; j++)
            {
                if(j == 1) printf("%d",Map[i][j]);
                else printf(" %d",Map[i][j]);
            }
            printf("\n");
        }
    }
    return 0;
} 

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