zoukankan      html  css  js  c++  java
  • 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
    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 Data Structures     Basic Data 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                  

    UESTC 1852 Traveling Cellsperson - qhn999 - 码代码的猿猿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 ",2*m-2);
            }
            else
            {
                if(n%2==0||m%2==0)
                {
                    printf("%d ",n*m);
                }
                else
                {
                    printf("%d ",m*n+1);
                }
            }
        }
        puts("LOL");
        return 0;
    }
    * This source code was highlighted by YcdoiT. ( style: Codeblocks )
  • 相关阅读:
    SQL SERVER 运维日记-数据库备份
    SQL调优日记--并行等待的原理和问题排查
    数据库服务器构建和部署最佳实践
    实战案例--TEMPDB暴涨
    关于TRIM的优化技巧
    SQL SERVER 的前世今生--各版本功能对比
    Sleeping会话导致阻塞原理(下)
    SQL SERVER运维日记--收缩数据库
    SQLSERVER 运维日记-数据库状态
    EXPERT FOR SQL SERVER诊断系列--索引
  • 原文地址:https://www.cnblogs.com/CKboss/p/3350828.html
Copyright © 2011-2022 走看看