POJ - 1475 Pushing Boxes
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这题我是1A的(其实在POJ上交了无数次)
做的时候一开始还读错题了,冷静调样例时才发现要Push数最少时Walk最少
具体思路:BFS啦,只是状态好像要记很多多东西啦,什么人的位置,箱子的位置啦,推了几次啊,走了几步啊
用个优先队列好像会快很多
在加个最优性剪枝好像就更快了(不加就死循环了)
然后就没啥了
花絮:震惊! 某学生被续走两天的时间! 原因竟是........优先队列没有清空辣鸡多组数据
AC代码
#include<cstdio> #include<cctype> #include<cstring> #include<queue> using namespace std; int n,m,t; char mp[100][100]; const int dir[4][2]={{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; const char Dir[4] = {\'S\',\'N\',\'E\',\'W\'}; struct point{int x,y;}; struct State{point box,man;int Push,Walk;}; int best[25][25][25][25][2]; bool operator <(State a,State b){if(a.Push!=b.Push)return a.Push>b.Push;else return a.Walk>b.Walk;} inline void BecomeBest(const State& state){best[state.box.x][state.box.y][state.man.x][state.man.y][0]=state.Push,best[state.box.x][state.box.y][state.man.x][state.man.y][1]=state.Walk;} priority_queue<State> q; inline bool Batter(const State& state){if(best[state.box.x][state.box.y][state.man.x][state.man.y][0]!=state.Push) return best[state.box.x][state.box.y][state.man.x][state.man.y][0]>state.Push;return best[state.box.x][state.box.y][state.man.x][state.man.y][1]>state.Walk;} inline bool isUreachable(point a){return a.x<=0||a.x>n||a.y<=0||a.y>m||mp[a.x][a.y]==\'#\';} void printPath(point b,point m,int p,int w) { if(p==0&&w==0)return; for(int i = 0; i < 4;i++) { point pm,pb; pm.x=m.x+dir[i][0],pm.y=m.y+dir[i][1]; if(isUreachable(pm))continue; if(!(pm.x==b.x&&pm.y==b.y)&&best[b.x][b.y][pm.x][pm.y][0]==p&&best[b.x][b.y][pm.x][pm.y][1]==w-1) { printPath(b,pm,p,w-1),putchar(tolower(Dir[i])); break; } pb.x=b.x+dir[i][0],pb.y=b.y+dir[i][1]; if(isUreachable(pb))continue; if(pb.x==m.x&&pb.y==m.y&&best[pb.x][pb.y][pm.x][pm.y][0]==p-1&&best[pb.x][pb.y][pm.x][pm.y][1]==w) { printPath(pb,pm,p-1,w),putchar(Dir[i]); break; } } } void solve() { while (!q.empty())q.pop(); int sx,sy,tx,ty,bx,by; memset(best,0x6F,sizeof(best)); for (int i=1; i<=n; i++) for (int j=1; j<=m; j++) if(mp[i][j]==\'S\')sx=i,sy=j; else if(mp[i][j]==\'B\')bx=i,by=j; State now,nex; now.box.x=bx,now.box.y=by,now.man.x=sx,now.man.y=sy,now.Push=0,now.Walk=0,q.push(now),BecomeBest(now); while(!q.empty()) { now=q.top(),q.pop(); if(mp[now.box.x][now.box.y] == \'T\') { printPath(now.box,now.man,now.Push,now.Walk),puts(""); return; } for(int i = 0; i < 4; ++i) { nex=now,nex.man.x+=dir[i][0],nex.man.y+=dir[i][1]; if(isUreachable(nex.man))continue; if(nex.man.x==now.box.x&&nex.man.y==now.box.y) { nex.box.x+=dir[i][0],nex.box.y+=dir[i][1]; if(isUreachable(nex.box))continue; nex.Push++; if(!Batter(nex))continue; q.push(nex),BecomeBest(nex); } else { nex.Walk++; if(!Batter(nex))continue; q.push(nex),BecomeBest(nex); } } } puts("Impossible."); } int main() { while (scanf("%d%d",&n,&m)) { if(n==0&&m==0)return 0; while(getchar() != \'\\n\'); for (int i=1; i<=n; i++)gets(mp[i]+1); printf("Maze #%d\\n",++t); solve(); puts(""); } return 0; }
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