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

Prev  Submit  Status  Statistics  Discuss  Next

Font Size:  +  -

Type:  None Graph Theory     2-SAT     Articulation/Bridge/Biconnected Component     Cycles/Topological Sorting/Strongly Connected Component     Shortest Path         Bellman Ford         Dijkstra/Floyd Warshall     Euler Trail/Circuit     Heavy-Light Decomposition     Minimum Spanning Tree     Stable Marriage Problem     Trees     Directed Minimum Spanning Tree     Flow/Matching         Graph Matching             Bipartite Matching             Hopcroft–Karp Bipartite Matching             Weighted Bipartite Matching/Hungarian Algorithm         Flow             Max Flow/Min Cut             Min Cost Max Flow DFS-like     Backtracking with Pruning/Branch and Bound     Basic Recursion     IDA* Search     Parsing/Grammar     Breadth First Search/Depth First Search     Advanced Search Techniques         Binary Search/Bisection         Ternary Search Geometry     Basic Geometry     Computational Geometry     Convex Hull     Pick's Theorem Game Theory     Green Hackenbush/Colon Principle/Fusion Principle     Nim     Sprague-Grundy Number Matrix     Gaussian Elimination     Matrix Exponentiation Da ta Structures     Basic Da ta Structures     Binary Indexed Tree     Binary Search Tree     Hashing     Orthogonal Range Search     Range Minimum Query/Lowest Common Ancestor     Segment Tree/Interval Tree     Trie Tree     Sorting     Disjoint Set String     Aho Corasick     Knuth-Morris-Pratt     Suffix Array/Suffix Tree Math     Basic Math     Big Integer Arithmetic     Number Theory         Chinese Remainder Theorem         Extended Euclid         Inclusion/Exclusion         Modular Arithmetic     Combinatorics         Group Theory/Burnside's lemma         Counting     Probability/Expected Value Others     Tricky     Hardest     Unusual     Brute Force     Implementation     Constructive Algorithms     Two Pointer     Bitmask     Beginner     Discrete Logarithm/Shank's Baby-step Giant-step Algorithm     Greedy     Divide and Conquer Dynamic Programming                     Tag it!

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 )