hdu 5336 XYZ and Drops

暴力模拟每个水滴和小水滴的状态即可，复杂度O（n*T）。

（大家可以想一下1<= r <= 100000,1 <= c <= 100000, 1<= n <= 100000, 1<= T <= 1000000000的时候怎么做）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
using namespace std;const int maxn=110;
struct node
{int x,y,val,t;
}no[maxn];
struct water
{int x,y,t,id,dir;water(){}water(int _x,int _y,int _t,int _id,int _dir):x(_x),y(_y),t(_t),id(_id),dir(_dir){}bool operator < (const water &a) const{return t > a.t;}
};int r,c,n,T;
int vis[maxn][maxn];void solve(int x,int y,priority_queue<water> &q,int t)
{int tmp = x;vis[x][y] = -1;while(tmp>0 && vis[tmp][y]==-1)tmp--;q.push(water(tmp,y,t+x-tmp,tmp==0?-1:vis[tmp][y],1));tmp = x;while(tmp<=r && vis[tmp][y]==-1)tmp++;q.push(water(tmp,y,t+tmp-x,tmp>r?-1:vis[tmp][y],2));tmp = y;while(tmp>0 && vis[x][tmp]==-1)tmp--;q.push(water(x,tmp,t+y-tmp,tmp==0?-1:vis[x][tmp],3));tmp = y;while(tmp<=c && vis[x][tmp]==-1)tmp++;q.push(water(x,tmp,t+tmp-y,tmp>c?-1:vis[x][tmp],4));
}
void bfs(int x,int y)
{priority_queue<water> q;solve(x,y,q,0);while(!q.empty()){water u = q.top();q.pop();if(u.id == -1)continue;if(u.t > T)continue;if(no[u.id].val <= 4 && vis[u.x][u.y] != -1){no[u.id].t = u.t;no[u.id].val++;if(no[u.id].val>4)solve(no[u.id].x,no[u.id].y,q,u.t);}else if(no[u.id].val>4 && no[u.id].t<u.t){int x = u.x,y = u.y;int tmp,t = u.t;if(u.dir == 1){tmp = x;while(tmp>0 && vis[tmp][y]==-1)tmp--;if(tmp > 0)q.push(water(tmp,y,t+x-tmp,vis[tmp][y],1));}else if(u.dir == 2){tmp = x;while(tmp<=r && vis[tmp][y]==-1)tmp++;if(tmp <= r)q.push(water(tmp,y,t+tmp-x,vis[tmp][y],2));}else if(u.dir == 3){tmp = y;while(tmp>0 && vis[x][tmp]==-1)tmp--;if(tmp > 0)q.push(water(x,tmp,t+y-tmp,vis[x][tmp],3));}else{tmp = y;while(tmp<=c && vis[x][tmp]==-1)tmp++;if(tmp <= c)q.push(water(x,tmp,t+tmp-y,vis[x][tmp],4));}}}
}int main()
{while(scanf("%d%d%d%d",&r,&c,&n,&T)!=EOF){memset(vis,-1,sizeof(vis));for(int i = 1;i <= n; i++){scanf("%d%d%d",&no[i].x,&no[i].y,&no[i].val);vis[no[i].x][no[i].y] = i;}int x,y;scanf("%d%d",&x,&y);bfs(x,y);for(int i = 1; i <= n; i++){if(no[i].val > 4)printf("0 %d\n",no[i].t);elseprintf("1 %d\n",no[i].val);}}return 0;
}

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