ACM: 广搜 图论题 poj 1708

                                            Game

 

Description

A child has drawn N(N<=100) numbered circles of different colors. Hehas connected some of the circles by colored oriented linesegments. Every pair of circles may have any number of segments,with any colors, connecting them. Each color (either circle coloror segment color) is assigned its own unique positive integernumber not greater than 100.

Starting the game the child first of all chooses three differentintegers L, K and Q within the range between 1 and N. Then heplaces one pawn into the circle number L and another one into thecircle number K, whereupon he begins to move them using thefollowing rules:

• a pawn can be moved along the line segment, if this segment hasthe same color with the circle where another pawn isplaced,
  
• the pawn can be moved only in the direction of the segment (allsegments are oriented),
  
• two pawns can never be placed in the same circle at the sametime,
  
• the move order is free (i.e. it is not necessary to move pawnsalternately),
  
• the game ends, when one of the pawns (any of the two) reachesthe last circle number Q.

You are to write a program to find out the shortest (i.e.containing a minimal number of moves) solution for this game, if itexists.

Input

The first line ofthe input file contains integers N, L, K, Q separated by spaces.The second line consists of N integers c1, c2, ... , cn, separatedby spaces, in the given order, where ci is the color of the circlenumber i. The third line consists of a single integer M(0<=M<=10000) denoting the totalnumber of segments. Then follow M lines, each containing adescription of one oriented segment. Each segment is described bythree integer numbers Aj, Bj, Cj, separated by spaces, where A andB are the numbers of the circles connected by the j-th segment withdirection from Aj to Bj, and Cj represents the color of thissegment.

Output

The first line ofthe output file should contain the word "YES", if the game can cometo the end, and "NO" otherwise (without quotes). If the answer is"YES", the second line of the output should contain just a singleinteger - the minimum number of the moves the child should make tofinish the game.

Sample Input

5 3 4 1
2 3 2 1 4
8
2 1 2
4 1 5
4 5 2
5 1 3
3 2 2
3 2 4
5 3 1
3 5 1

Sample Output

YES
3

题意: 现在有两个点, L,K为两个起点编号, 计算最少的移动次数到达Q点,若可以就输出YES和最少步数,

     否则输出NO.

     但是没移动一次有要求:

                        1. 只可以沿着线段方向移动.

                         2.两个人不可以同时移动到同一个点上.

                        3. 移动次序是随意的.

                        4. 谁先到达就结束游戏.

解题思路:

         1. 广搜第一反应. 也是正确的.

         2. 怎么广搜呢? 用什么数据类型? 现在开始会思考这些问题. (进步了！！)

          3.设每个队列的节点同时存在L，K的当前位置, 即使假设他们同步进行行走, 模拟过程.

         4. 主要是要记得判断他们不可以同时移动到同一个点上.

 

代码:

#include<cstdio>
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
#define MAX 105
const int INF = (1<<29);

struct node
{
 int v;
 int color;
 int next;
}edges[MAX*MAX];

struct node1
{
 int posL, posK;
 int step;
};

int N, L, K, Q, m;
int first[MAX], num;
bool vis[MAX][MAX];
int color[MAX];
int result;

inline void add(int u,int v,int co)
{
 edges[num].v = v;
 edges[num].color = co;
 edges[num].next = first[u];
 first[u] = num++;
}

void read_graph()
{
 memset(vis,false,sizeof(vis));
 memset(first,-1,sizeof(first));
 memset(edges,0,sizeof(edges));
 num = 0;
 scanf("%d",&m);
 int u, v, co;
 int i;
 for(i = 1; i <= m; ++i)
 {
  scanf("%d %d%d",&u,&v,&co);
  add(u,v,co);
 }
}

bool bfs()
{
 queue<node1>qu;
 node1 temp;
 temp.posK = K, temp.posL = L, temp.step =0;
 vis[K][L] = true;
 qu.push(temp);
 bool flag = false;
 int e;
 result = INF;
 
 while( !qu.empty() )
 {
  temp = qu.front();
  qu.pop();
  if(temp.posK == Q || temp.posL== Q)
  {
   if(temp.step< result) result = temp.step;
   flag =true;
  }

  for(e =first[temp.posL]; e != -1; e = edges[e].next)
  {
   if(!vis[edges[e].v ][ temp.posK ] &&edges[e].v != temp.posK &&edges[e].color == color[temp.posK])
   {
    vis[edges[e].v ][ temp.posK ] = true;
    node1t;
    t.posK= temp.posK;
    t.posL= edges[e].v;
    t.step= temp.step+1;
    qu.push(t);
   }
  }

  for(e =first[temp.posK]; e != -1; e = edges[e].next)
  {
   if( !vis[temp.posL ][ edges[e].v ] &&edges[e].v != temp.posL &&edges[e].color == color[temp.posL] )
   {
    vis[temp.posL ][ edges[e].v ] = true;
    node1t;
    t.posK= edges[e].v;
    t.posL= temp.posL;
    t.step= temp.step+1;
    qu.push(t);
   }
  }
 }

 return flag;
}

int main()
{
 int i;
// freopen("input.txt","r",stdin);
 while(scanf("%d %d %d%d",&N,&L,&K,&Q)!= EOF)
 {
  for(i = 1; i <=N; ++i)
   scanf("%d",&color[i]);
  read_graph();
  if( !bfs() )
   printf("NO\n");
  else
   printf("YES\n%d\n",result);
 }

 return 0;
}