这是一道很简单的图论题,只要使用宽度优先搜索(BFS)标记节点间距离即可。

我的解题代码如下:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <string>
#include <algorithm> #include <queue>
using namespace std; int adj[20][20];
int dis[20]; //为正时表示各点到s的距离,为-1时表示该点还未被bfs遇到
int SP(int s,int t)
{//使用BFS,返回t到s的最短距离
memset(dis,-1,sizeof(dis));
queue<int> q;
dis[s]=0;
q.push(s);
while(!q.empty())
{
int u=q.front(); q.pop();
if(u==t) break;
for(int j=0; j<20; j++) if(dis[j]<0 && adj[u][j])
{
q.push(j);
dis[j]=dis[u]+1; //j到s的距离为u到s的距离+1
}
}
return dis[t];
}
int main()
{
int X,T=0;
while(scanf("%d",&X)==1)
{
int tmp;
memset(adj,0,sizeof(adj));
for(int j=0; j<X; j++)
{
scanf("%d",&tmp); adj[0][tmp-1]=adj[tmp-1][0]=1;
}
for(int i=1; i<19; i++)
{
scanf("%d",&X);
for(int j=0; j<X; j++)
{
scanf("%d",&tmp); adj[i][tmp-1]=adj[tmp-1][i]=1;
}
}
scanf("%d",&tmp);
printf("Test Set #%d\n",++T);
int A,B;
for(int i=0; i<tmp; i++)
{
scanf("%d %d",&A,&B);
printf("%2d to %2d: %d\n",A,B,SP(A-1,B-1));
}
printf("\n");
}
return 0;
}

附上题目如下:

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 
I
th 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 (


) indicating the number of country pairs that follow. The next 
N
 lines each contain exactly two integers (


) 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 countryB 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

UVa 567: Risk的更多相关文章

  1. UVA 567 Risk【floyd】

    题目链接: option=com_onlinejudge&Itemid=8&page=show_problem&problem=508">https://uva ...

  2. uva oj 567 - Risk(Floyd算法)

    /* 一张有20个顶点的图上. 依次输入每个点与哪些点直接相连. 并且多次询问两点间,最短需要经过几条路才能从一点到达另一点. bfs 水过 */ #include<iostream> # ...

  3. uva 12264 Risk

    https://vjudge.net/problem/UVA-12264 题意: 有很多个阵地,分为敌方和己方,每个士兵可以移动到相邻的己方的阵地,但是只能移动一步. 现在要让与敌方相邻的阵地中士兵最 ...

  4. uva 567

    Floyd 算法   就输入麻烦点 #include <iostream> #include <cstring> #include <cstdlib> #inclu ...

  5. UVA - 12264 Risk (二分,网络流)

    题意比较坑,移动完以后的士兵不能再次移动,不然样例都过不了... 最小值最大满足决策单调性所以二分答案,跑网络流验证是否可行. 这种题重点在建图,为了保证只移动一次,拆点,一个入点一个出点,到了出点的 ...

  6. UVA题目分类

    题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics ...

  7. Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) Volume 7. Graph Algorithms and Implementation Techniques

    uva 10803 计算从任何一个点到图中的另一个点经历的途中必须每隔10千米 都必须有一个点然后就这样 floy 及解决了 ************************************* ...

  8. Risk UVA - 12264 拆点法+最大流+二分 最少流量的节点流量尽量多。

    /** 题目:Risk UVA - 12264 链接:https://vjudge.net/problem/UVA-12264 题意:给n个点的无权无向图(n<=100),每个点有一个非负数ai ...

  9. UVA - 1025 A Spy in the Metro[DP DAG]

    UVA - 1025 A Spy in the Metro Secret agent Maria was sent to Algorithms City to carry out an especia ...

随机推荐

  1. POJ 1716 Integer Intervals 差分约束

    题目:http://poj.org/problem?id=1716 #include <stdio.h> #include <string.h> #include <ve ...

  2. Web负载均衡的几种方式

    Web负载均衡的几种实现方式 摘要:负载均衡(Load Balance)是集群技术(Cluster)的一种应用.负载均衡可以将工作任务分摊到多个处理单元,从而提高并发处理能力.目前最常见的负载均衡应用 ...

  3. VS2010升级VS2012必备(MVC4 WebPage2.0 Razor2.0资料汇集)

    刚把项目升级到2012,发现发生了很多变化,以下是最近看过的网站和资料汇集,供需要者参考. 本文在最近一个月可能会不断更新. Razor2.0 新特性介绍: 介绍1:http://vibrantcod ...

  4. 7 -- Spring的基本用法 -- 8...

    7.8 深入理解容器中的Bean 7.8.1 抽象Bean与子Bean 把多个<bean.../>配置中相同的信息提取出来,集中成配置模版------这个配置模版并不是真正的Bean,因此 ...

  5. Web API路由与动作(三)

    本章包括三个小节  如果你输入了mvc的路由规则 这个可以粗略过一遍即可  内容说明有点繁琐 原文地址:http://www.asp.net/web-api/overview/web-api-rout ...

  6. Android事件分发详解(三)——ViewGroup的dispatchTouchEvent()源码学习

    package cc.aa; import android.os.Environment; import android.view.MotionEvent; import android.view.V ...

  7. vi的撤销命令

    'u' : 撤销上一个编辑操作 'ctrl + r' : 恢复,即回退前一个命令 'U' : 行撤销,撤销所有在前一个编辑行上的操作 使用u 和 CTRL+R 命令可以恢复到任何编辑过的状态

  8. javascript循环

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...

  9. POJ_2184_Cow_Exhibition_(动态规划,背包)

    描述 http://poj.org/problem?id=2184 n只奶牛,每只都有智商s_i和情商f_i,取出若干只,保证智商之和与情商之和都不为负的情况下,让两者之和最大. Cow Exhibi ...

  10. 【转】Fragment和Activity

    原文网址:http://www.cnblogs.com/mengdd/archive/2013/01/11/2856374.html Fragment和Activity的交互 一个Fragment的实 ...