zoukankan      html  css  js  c++  java
  • uva 567

     

      Risk 

    Risk is a board game in which several opposing players attempt to conquer the world. The gameboard consists of a world map broken up into hypothetical countries. During a player's turn, armies stationed in one country are only allowed to attack only countries with which they share a common border. Upon conquest of that country, the armies may move into the newly conquered country.


    During the course of play, a player often engages in a sequence of conquests with the goal of transferring a large mass of armies from some starting country to a destination country. Typically, one chooses the intervening countries so as to minimize the total number of countries that need to be conquered. Given a description of the gameboard with 20 countries each with between 1 and 19 connections to other countries, your task is to write a function that takes a starting country and a destination country and computes the minimum number of countries that must be conquered to reach the destination. You do not need to output the sequence of countries, just the number of countries to be conquered including the destination. For example, if starting and destination countries are neighbors, then your program should return one.


    The following connection diagram illustrates the first sample input.

    Input 

    Input to your program will consist of a series of country configuration test sets. Each test set will consist of a board description on lines 1 through 19. The representation avoids listing every national boundary twice by only listing the fact that country I borders country J when I < J. Thus, the Ith line, where I is less than 20, contains an integer X indicating how many ``higher-numbered" countries share borders with country I, then X distinct integers J greater than I and not exceeding 20, each describing a boundary between countries I and J. Line 20 of the test set contains a single integer ( $1 le N le 100$) indicating the number of country pairs that follow. The next N lines each contain exactly two integers ( $1 le A,B le 20; A 
e B$) indicating the starting and ending countries for a possible conquest.


    There can be multiple test sets in the input file; your program should continue reading and processing until reaching the end of file. There will be at least one path between any two given countries in every country configuration.

    Output 

    For each input set, your program should print the following message ``Test Set #T" where T is the number of the test set starting with 1 (left-justified starting in column 11).

    The next NT lines each will contain the result for the corresponding test in the test set - that is, the minimum number of countries to conquer. The test result line should contain the start country code A right-justified in columns 1 and 2; the string `` to " in columns 3 to 6; the destination country code B right-justified in columns 7 and 8; the string ``: " in columns 9 and 10; and a single integer indicating the minimum number of moves required to traverse from country A to country B in the test set left-justified starting in column 11. Following all result lines of each input set, your program should print a single blank line.

    Sample Input 

    1 3
    2 3 4
    3 4 5 6
    1 6
    1 7
    2 12 13
    1 8
    2 9 10
    1 11
    1 11
    2 12 17
    1 14
    2 14 15
    2 15 16
    1 16
    1 19
    2 18 19
    1 20
    1 20
    5
    1 20
    2 9
    19 5
    18 19
    16 20
    4 2 3 5 6
    1 4
    3 4 10 5
    5 10 11 12 19 18
    2 6 7
    2 7 8
    2 9 10
    1 9
    1 10
    2 11 14
    3 12 13 14
    3 18 17 13
    4 14 15 16 17
    0
    0
    0
    2 18 20
    1 19
    1 20
    6
    1 20
    8 20
    15 16
    11 4
    7 13
    2 16
    

    Sample Output 

    Test Set #1
     1 to 20: 7
     2 to  9: 5
    19 to  5: 6
    18 to 19: 2
    16 to 20: 2
    
    Test Set #2
     1 to 20: 4
     8 to 20: 5
    15 to 16: 2
    11 to  4: 1
     7 to 13: 3
     2 to 16: 4
    #include <iostream>
    #include <stack>
    #include <cstring>
    #include <cstdio>
    #include <string>
    #include <algorithm>
    #include <queue>
    #include <set>
    #include <map>
    #include <fstream>
    #include <stack>
    #include <list>
    #include <sstream>
    #include <cmath>
    
    using namespace std;
    
    #define ms(arr, val) memset(arr, val, sizeof(arr))
    #define mc(dest, src) memcpy(dest, src, sizeof(src))
    #define N 25
    #define INF 0x3fffffff
    #define vint vector<int>
    #define setint set<int>
    #define mint map<int, int>
    #define lint list<int>
    #define sch stack<char>
    #define qch queue<char>
    #define sint stack<int>
    #define qint queue<int>
    /*弗洛伊德算法*/
    int g[N][N];
    
    void floyd()
    {
        for (int k = 1; k <= 20; k++)
        {
            for (int i = 1; i <= 20; i++)
            {
                for (int j = 1; j <= 20; j++)
                {
                    if (g[i][j] > g[i][k] + g[k][j])
                    {
                        g[i][j] = g[i][k] + g[k][j];
                    }
                }
            }
        }
    }
    
    int main()
    {
        int n, cs = 1, t;
        while (~scanf("%d", &n))
        {
            for (int i = 1; i <= 20; i++)
            {
                for (int j = 1; j <= 20; j++)
                {
                    g[i][j] = INF;
                }
            }
            for (int i = 0; i < n; i++)
            {
                scanf("%d", &t);
                g[1][t] = g[t][1] = 1;
            }
    
            for (int i = 2; i <= 19; i++)
            {
                scanf("%d", &n);
                for (int j = 0; j < n; j++)
                {
                    scanf("%d", &t);
                    g[i][t] = g[t][i] = 1;
                }
            }
            floyd();
            printf("Test Set #%d
    ", cs++);
            scanf("%d", &n);
            int s, e;
            while (n--)
            {
                scanf("%d%d", &s, &e);
                printf("%2d to %2d: %d
    ", s, e, g[s][e]);
            }
            putchar('
    ');
        }
        return 0;
    }


  • 相关阅读:
    使用VS2005搭建典型高效的SharePoint开发环境,提高生产效率,包含远程调试,自动部署 无为而为
    该死的Windows 2003 Server DMA设置,然我刻录DVD这么慢,终于找到办法了 无为而为
    听说Team Foundation Server 繁體中文版於 2006.04.28 RTM ,不知道微软中国的工作做得如何? 无为而为
    [软件开发过程]反模式:简单的部分留在需求人员的脑海中,只描述最复杂的部分给我们听 无为而为
    软件过程改进(SPI)常见反模式:第22条军规 无为而为
    在WebPart中上传图片到SharePoint图片库,读取Exif信息到图片的自定义属性 无为而为
    SQL Server 2005 Service Pack 1正式发布了,我想起,有人说,微软的软件至少要等到出了SP1才能用,那么现在可以用SQL2005了 无为而为
    可以下载Microsoft ISA Server 2006 试用版了,网管需要关注 无为而为
    超女带给我们什么? 无为而为
    Asp.net Url改写方法——采用HttpModules(转)
  • 原文地址:https://www.cnblogs.com/jecyhw/p/3916872.html
Copyright © 2011-2022 走看看