Risk（最短路径Floyd算法）

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 <= N <= 100) indicating the number of country pairs that follow. The next N lines each contain exactly two integers (1 <= A,B <= 20; A!=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
3 7 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

一开始理解题目花了很多时间，题目是给出一个类似于邻接表的东西，19行是确定的，如果没有联通就是0，之后就是Floyd算法的运用，迪杰斯特里拉算法才刚学会，还是靠的百度模板才成功ac的题目，有不懂的可以去这里看一下，我觉得写的很好。http://www.cnblogs.com/biyeymyhjob/archive/2012/07/31/2615833.html。

这道题目要注意的就是输入输出的问题，其他应该就是算法的实现了。

Floyd算法的核心：从任意节点i到任意节点j的最短路径不外乎2种可能，1是直接从i到j，2是从i经过若干个节点k到j。
所以，我们假设Dis(i,j)为节点u到节点v的最短路径的距离，对于每一个节点k，
我们检查Dis(i,k) + Dis(k,j) < Dis(i,j)是否成立，如果成立，证明从i到k再到j的路径比i直接到j的路径短，
我们便设置Dis(i,j) = Dis(i,k) + Dis(k,j)，这样一来，当我们遍历完所有节点k，Dis(i,j)中记录的便是i到j的最短路径的距离。

#include<iostream>  #include<cstdio>  const int MAXN=25;  using namespace std;  int map[MAXN][MAXN];  void floyd(int n) {      int i,j,k;      for(k=1;k<=n;k++)          for(i=1;i<=n;i++)              for(j=1;j<=n;j++)                  if(map[i][k]+map[k][j]<map[i][j])                      map[i][j]=map[i][k]+map[k][j];  }  int main()  {        int n,x,i,start,end,csc=0;      while(~scanf("%d",&n))      {          memset(map,63,sizeof(map));          while(n--)          {              scanf("%d",&x);              map[1][x]=map[x][1]=1;          }          for(i=2;i<=19;i++)          {              scanf("%d",&n);              while(n--)              {                  scanf("%d",&x);                  map[i][x]=map[x][i]=1;              }          }          floyd(20);          scanf("%d",&n);          printf("Test Set #%d\n",++csc);          while(n--)          {              scanf("%d%d",&start,&end);              printf("%d to %d: %d\n",start,end,map[start][end]);            }          printf("\n");      }  }