TOJ 2755 国际象棋（搜索）

传送门:http://acm.tzc.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=2755

思路：对起点到终点进行广搜,由于n特别大，不能用二维数组记录走过的点，可以用STL的map进行记录，map<pair<int,int>,int> v; 如果出现的点可以进行v[point] = 1；

另外可以剪枝  if(abs(sx-ex)/2>m||abs(sy-ey)/2>m)return 0;  其中sx,sy代表起点的横纵坐标，ex,ey代表终点的横纵坐标

AC代码如下：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<cmath>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<vector>
#define LL long long
#include<assert.h>
using namespace std;
int go[][]={{,},{,-},{-,},{-,-},{,},{,-},{-,},{-,-}};
int n;
map<pair<int,int>,int> v;
struct note{
    int x,y,step;
}pos,q;
int check(int x,int y){
    if(x <  ||x > n || y <  || y > n)return ;
    return ;
}
int bfs(int sx,int sy,int ex,int ey,int m){
    pair<int ,int> a;
    a.first = sx;a.second = sy;
    v[a] = ;
    if(abs(sx-ex)/>m||abs(sy-ey)/>m)return ;//剪枝
    queue<note>que;
    while(!que.empty()) que.pop();

    pos.x = sx;pos.y = sy;pos.step = ;
    que.push(pos);
    while(que.size()){
        q = que.front();
        que.pop();
        if(q.step > m)return ;
        if(q.x == ex && q.y == ey && q.step <= m) return ;
        for(int i =  ; i <  ; i ++){
            int dx = q.x + go[i][];
            int dy = q.y + go[i][];
            pair<int,int>b;
            b.first = dx;b.second = dy;
            if(check(dx,dy) && q.step < m && v[b] == ){
                pos.step = q.step + ;
                pos.x = dx;pos.y = dy;
                v[b] = ;
                if(dx == ex && dy == ey && pos.step <= m) return ;
                que.push(pos);
            }
        }
    }
    return ;
}
int main(){
    int m;
    while(~scanf("%d %d",&n,&m)){
         v.clear();
         int sx,sy,ex,ey;
         scanf("%d %d %d %d",&sx,&sy,&ex,&ey);
         if(!bfs(sx,sy,ex,ey,m)){
             printf("Knight cannot reach Queen within %d moves!\n",m);
         }
         else{
             printf("Knight can reach Queen within %d moves!\n",m);
         }

    }
}