poj 1915 Knight Moves 【双向bfs】

标签：双向bfs

Description

Background 
Mr Somurolov, fabulous chess-gamer indeed, asserts that no one else but him can move knights from one position to another so fast. Can you beat him? 
The Problem 
Your task is to write a program to calculate the minimum number of moves needed for a knight to reach one point from another, so that you have the chance to be faster than Somurolov. 
For people not familiar with chess, the possible knight moves are shown in Figure 1. 

Input

Output

Sample Input

3
8
0 0
7 0
100
0 0
30 50
10
1 1
1 1

Sample Output

5
28
0

分析：最基础的bfs，没用单向的广搜，学习了一下怎么用双向的广搜。

以下仅个人观点。。。

能用到双向广搜的题，首先得有开始点和目标点，然后就是从开始和目标同时出发，分别进行单向搜索，知道某一方向的搜索碰到另一方向搜索过得点为止。。

#include <stdio.h>
#include <queue>
#include <string.h>
#define  M 305
using namespace std;

int dis1[M][M], dis2[M][M], len;
const int dx[] = {1, 1, -1, -1, 2, 2, -2, -2};  //方向
const int dy[] = {2, -2, 2, -2, 1, -1, 1, -1};
struct node{
    int x, y;
}st, en;


int limit(node a){
    return (a.x>=0&&a.x<len&&a.y>=0&&a.y<len);
}

void dobfs(){
   int i, size;
   dis1[st.x][st.y] = dis2[en.x][en.y] = 0;
   queue<node> p, q;
   q.push(st); p.push(en);
   while(!q.empty()&&!p.empty()){
        size = q.size();
        while(size --){  
            node cur = q.front();
            q.pop();
            if(limit(cur)&&dis2[cur.x][cur.y] != -1){
                //printf("%d %d..1\n", dis1[cur.x][cur.y], dis2[cur.x][cur.y]);
                printf("%d\n", dis1[cur.x][cur.y]+dis2[cur.x][cur.y]);
                return ;
            }
            for(i = 0; i < 8; i ++){
                node temp = cur;
                temp.x+=dx[i], temp.y += dy[i];
                if(limit(temp)&&dis2[temp.x][temp.y] != -1){
                     //printf("%d %d....1\n", dis1[cur.x][cur.y], dis2[temp.x][temp.y]);
                    printf("%d\n", dis1[cur.x][cur.y]+dis2[temp.x][temp.y]+1);
                    return ;
                }
                if(limit(temp)&&dis1[temp.x][temp.y] == -1){
                    q.push(temp); dis1[temp.x][temp.y] = dis1[cur.x][cur.y]+1;
                }
            }
        }
        size = p.size();
        while(size --){
            node cur = p.front();
            p.pop();
            if(limit(cur)&&dis1[cur.x][cur.y] != -1){
                 //printf("%d %d..2\n", dis1[cur.x][cur.y], dis2[cur.x][cur.y]);
                printf("%d\n", dis1[cur.x][cur.x]+dis2[cur.x][cur.y]);
                return ;
            }
            for(i = 0; i < 8; i ++){
                node temp = cur;
                temp.x += dx[i]; temp.y += dy[i];
                if(limit(temp)&&dis1[temp.x][temp.y] != -1){
                     //printf("%d %d....2\n", dis2[cur.x][cur.y], dis1[temp.x][temp.y]);
                    printf("%d\n", dis1[temp.x][temp.y]+dis2[cur.x][cur.y]+1);
                    return ;
                }
                if(limit(temp)&&dis2[temp.x][temp.y] == -1){
                    p.push(temp); dis2[temp.x][temp.y] = dis2[cur.x][cur.y] + 1;
                }
            }
        }
   }
}

int main(){
    int t;
    scanf("%d", &t);
    while(t --){
        scanf("%d", &len);
        scanf("%d%d%d%d", &st.x, &st.y, &en.x, &en.y);
        memset(dis1, -1, sizeof(dis1));
        memset(dis2, -1, sizeof(dis2));
        if(match(st, en)){
            printf("0\n"); continue;
        }
        else   dobfs();
    }
    return 0;
}

poj 1915 Knight Moves 【双向bfs】

标签：双向bfs

原文地址：http://blog.csdn.net/shengweisong/article/details/40343477