UESTC 1852 Traveling Cellsperson

找规律水题。。。

Traveling Cellsperson

Time Limit: 1000ms

Memory Limit: 65535KB

This problem will be judged on UESTC. Original ID: 1852
64-bit integer IO format: %lld      Java class name: Main

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You have solved every problem from Project Euler in your head. Now it is time for a problem you might have heard of,namely The Traveling Salesperson, whose decision version is NP-complete. We consider the Traveling Salesperson problem in a 2D rectangular grid where every cell can be reached from their neighboring cells (up,down, left and right) and you can visit a cell as many times as you like (though, most of the cells aren't that interesting, so you might prefer not to visit them a lot).

Input

The first line of the input consists of a single integer T, the number of test cases. Then follow two integers X and Y , marking the width and height of the grid, respectively. Then follow Y lines with X characters, where the character 'C' is a cell and the character 'S' is the starting point.

0 < T <= 50
0 < X <= 100
0 < Y <= 100
All characters in a test case are 'C', except for exactly one, which is 'S'.

Output

For each test case, output the minimum number of steps required to make a full roundtrip of the grid, starting and ending at S, and visiting each cell at least once.
Since you realize that this won't lead anywhere, finish off the output with "LOL"(without quotes) on a line of its own (one per run, not per test case).

Sample Input

1
4 4
CCCC
CCCC
CSCC
CCCC

Sample Output

16
LOL

Source

IDI Open 2013 Programming Contest

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

char str[110];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=0;i<m;i++)
            scanf("%s",str);
        if(n>m) swap(n,m);
        if(n==1)
        {
            printf("%d\n",2*m-2);
        }
        else
        {
            if(n%2==0||m%2==0)
            {
                printf("%d\n",n*m);
            }
            else
            {
                printf("%d\n",m*n+1);
            }
        }
    }
    puts("LOL");
    return 0;
}

* This source code was highlighted by YcdoiT. ( style: Codeblocks )