Risk
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 2915   Accepted: 1352

Description

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. 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 the string " to " the destination country code B ; the string ": " and a single integer indicating the minimum number of moves required to traverse from country A to country B in the test set. 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

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
题解:很水的一道题,就是个最简单的最短路;
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
#define mem(x,y) memset(x,y,sizeof(x))
const int MAXN=;
int mp[MAXN][MAXN],d[MAXN],vis[MAXN];
void dijkscra(int u){
mem(vis,);
mem(d,INF);
d[u]=;
while(true){
int k=-;
for(int i=;i<=;i++){
if(!vis[i])if(k==-||d[i]<d[k])k=i;
}
if(k==-)break;
vis[k]=;
for(int i=;i<=;i++)
d[i]=min(d[i],d[k]+mp[k][i]);
}
}
int main(){
int kase=;
while(){
mem(mp,INF);
int x,b,a;
for(int i=;i<=;i++){
if(!~scanf("%d",&x))exit();
while(x--){
scanf("%d",&b);
mp[i][b]=mp[b][i]=;
}
}
if(!~scanf("%d",&x))exit();
printf("Test Set #%d\n",++kase);
while(x--){
scanf("%d%d",&a,&b);
dijkscra(a);
printf("%d to %d: %d\n",a,b,d[b]);
}
puts("");
}
return ;
}

Risk(最短路)的更多相关文章

  1. UVa567_Risk(最短路)(小白书图论专题)

    解题报告 option=com_onlinejudge&Itemid=8&category=7&page=show_problem&problem=508"& ...

  2. 任意两点之间的最短路(floyed)

    F.Moving On Firdaws and Fatinah are living in a country with nn cities, numbered from 11 to nn. Each ...

  3. bzoj1001--最大流转最短路

    http://www.lydsy.com/JudgeOnline/problem.php?id=1001 思路:这应该算是经典的最大流求最小割吧.不过题目中n,m<=1000,用最大流会TLE, ...

  4. 【USACO 3.2】Sweet Butter(最短路)

    题意 一个联通图里给定若干个点,求他们到某点距离之和的最小值. 题解 枚举到的某点,然后优先队列优化的dijkstra求最短路,把给定的点到其的最短路加起来,更新最小值.复杂度是\(O(NElogE) ...

  5. Sicily 1031: Campus (最短路)

    这是一道典型的最短路问题,直接用Dijkstra算法便可求解,主要是需要考虑输入的点是不是在已给出的地图中,具体看代码 #include<bits/stdc++.h> #define MA ...

  6. 最短路(Floyd)

    关于最短的先记下了 Floyd算法: 1.比较精简准确的关于Floyd思想的表达:从任意节点A到任意节点B的最短路径不外乎2种可能,1是直接从A到B,2是从A经过若干个节点X到B.所以,我们假设maz ...

  7. bzoj1266最短路+最小割

    本来写了spfa wa了 看到网上有人写Floyd过了 表示不开心 ̄へ ̄ 改成Floyd试试... 还是wa ヾ(。`Д´。)原来是建图错了(样例怎么过的) 结果T了 于是把Floyd改回spfa 还 ...

  8. HDU2433 BFS最短路

    Travel Time Limit: 10000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Sub ...

  9. 最短路(代码来源于kuangbin和百度)

    最短路 最短路有多种算法,常见的有一下几种:Dijstra.Floyd.Bellman-Ford,其中Dijstra和Bellman-Ford还有优化:Dijstra可以用优先队列(或者堆)优化,Be ...

随机推荐

  1. SIEM思考

    https://securosis.com/blog/comments/understanding-and-selecting-siem-log-management-introduction/ ht ...

  2. COB(Chip On Board)的製程簡單介紹

    前面提及 COB 的生產與 IC 的封裝製程幾乎是一致的,除了把 leadframe 改成了 PCB,把封膠由 molding 改成 dispensing,少了 triming & marki ...

  3. NFC 与点对点应用

    http://wenku.baidu.com/view/6a7623a28762caaedc33d426.html

  4. 充分利用CPU高速缓存,提高程序效率(原理篇)

    提高程序效率应该充分利用CPU的高速缓存.要想编写出对CPU缓存友好的程序就得先明白CPU高速缓存的运行机制. i5-2400S: 1.有三级缓存分别为 32k(数据.指令缓存分开,分为32k),25 ...

  5. MindManager_9.1.157使用模板时显示“参数错误”

    每次使用标准模板时都出现这个问题,上网搜索,原来是模板中存在的 “注释”导致的问题.具体原因不详,解决起来也简单,就是繁琐一点.转抄如下: 先找到模板文件夹,共四个文件夹Communcation.Pe ...

  6. http 中get和post

    1. http://www.cnblogs.com/hyddd/archive/2009/03/31/1426026.html 浅谈HTTP中Get与Post的区别 2009-03-31 14:51 ...

  7. 全国计算机等级考试二级教程-C语言程序设计_第3章_顺序结构

    1输入两个整数给变量x和y:然后输出x和y:在交换x和y中的值后,在输出x和y. #include <stdio.h> main() { int x, y, t; printf(" ...

  8. ARM上的linux如何实现无线网卡的冷插拔和热插拔

    ARM上的linux如何实现无线网卡的冷插拔和热插拔 fulinux 凌云实验室 1. 冷插拔 如果在系统上电之前就将RT2070/RT3070芯片的无线网卡(以下简称wlan)插上,即冷插拔.我们通 ...

  9. UVA11387 - The 3-Regular Graph(推理)

    题目链接 题意:给n个点,问能否画出一个无向图.且每一个顶点连接3条边.假设能够的话输出连接的边. 思路:当添加一条边时,总的无向图的度数会添加2,所以度数之和n*2为偶数.当n为奇数时,度数之和为奇 ...

  10. android典型监听事件实

    public class MainActivity extends Activity { int counter; Button add, sub; TextView display; @Overri ...