FZU2150 fire 双搜bfs

题目链接：http://acm.fzu.edu.cn/problem.php?pid=2150
题意：在任意两处点火，求最短时间烧光所有草堆。
规模：1 <= T <=100, 1 <= n <=10, 1 <= m <=10
类型：暴力+双搜
分析：题目很好理解，枚举两个起点进行双搜就好。但以前没有敲过双搜，中间出了不少状况，wa了很多发，主要问题是在“如何使两个queue完成当前时间下的搜索”。

时间复杂度&&优化：
代码：

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<stack>#include<math.h>#include<vector>#include<algorithm>#include<iostream>using namespace std;const int MAX_N = 100005;const int MAX_M = 5005;const int inf = 1000000007;const int mod = 1000000007;int n,m;int a[15][15];int b[15][15];struct point{    int x,y;    int step;};int dx[4]={0,0,1,-1};int dy[4]={1,-1,0,0};bool judge(int x,int y){    if(x>=0&&x<n&&y>=0&&y<m&&b[x][y]==1){            b[x][y]=0;        //cout<<"!"<<endl;        return true;    }    return false;}int bfs(int ax,int ay,int bx,int by){    int res=0;    for(int i=0;i<n;i++){        for(int j=0;j<m;j++){            b[i][j]=a[i][j];        }    }    b[ax][ay]=0;    b[bx][by]=0;    queue<point>q,p;    while(!q.empty()){q.pop();}    while(!p.empty()){p.pop();}    point st1,st2;    st1.x=ax;st1.y=ay;st1.step=0;    st2.x=bx;st2.y=by;st2.step=0;    q.push(st1);p.push(st2);    int stp=0;    point tmp,tmp2;    while(!q.empty() || !p.empty()){        if(!q.empty()){            int key_step=q.front().step;            while(!q.empty()&&q.front().step==key_step){                tmp=q.front();                //cout<<tmp.x<<" "<<tmp.y<<" "<<tmp.step<<endl;                q.pop();                stp=tmp.step;                res=max(res,stp);                for(int i=0;i<4;i++){                    tmp2.x=tmp.x+dx[i];                    tmp2.y=tmp.y+dy[i];                    tmp2.step=tmp.step+1;                    if(judge(tmp2.x,tmp2.y))q.push(tmp2);                }//                if(ax==1&&ay==1&&bx==7&&by==2){//                    cout<<stp<<" "<<tmp.x<<" "<<tmp.y<<endl;//                }            }        }        if(!p.empty()){            int key_step=p.front().step;            while(!p.empty()&&p.front().step==key_step){                tmp=p.front();                p.pop();                stp=tmp.step;                res=max(res,stp);                for(int i=0;i<4;i++){                    tmp2.x=tmp.x+dx[i];                    tmp2.y=tmp.y+dy[i];                    tmp2.step=tmp.step+1;                    if(judge(tmp2.x,tmp2.y))p.push(tmp2);                }//                if(ax==1&&ay==1&&bx==7&&by==2){//                    cout<<stp<<" "<<tmp.x<<" "<<tmp.y<<endl;//                }            }        }    }    for(int i=0;i<n;i++){        for(int j=0;j<m;j++){            if(b[i][j]==1)return -1;        }    }//    //cout<<stp<<" "<<ax<<" "<<ay<<"   "<<bx<<" "<<by<<endl;//    if(stp==3)cout<<3<<" "<<ax<<" "<<ay<<"   "<<bx<<" "<<by<<endl;//    if(stp==4)cout<<4<<" "<<ax<<" "<<ay<<"   "<<bx<<" "<<by<<endl;    return res;}int solve(){    int res=inf;    for(int i=0;i<n*m;i++){        if(a[i/m][i%m]==1)        for(int j=i+1;j<n*m;j++){            if(a[j/m][j%m]==1){               // cout<<i<<" "<<j<<endl;                int k=bfs(i/m,i%m,j/m,j%m);                if(k!=-1)res=min(res,k);            }        }    }    if(res==inf)return -1;    else return res;}int main(){    //freopen("test.txt","r",stdin);    int T;char c;    cin>>T;    for(int z=1;z<=T;z++){        memset(a,0,sizeof(a));        cin>>n>>m;        int tot=0;        for(int i=0;i<n;i++){            for(int j=0;j<m;j++){                cin>>c;                if(c=='#'){a[i][j]=1;tot++;}                else a[i][j]=0;            }        }        if(tot<=2){printf("Case %d: %d\n",z,0);}        else printf("Case %d: %d\n",z,solve());    }    return 0;}