XYZ and Drops
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 714 Accepted Submission(s): 196
In every second, every small drop moves to the next cell of its direction. It is possible that multiple small drops can be at same cell, and they won't collide. Then for each cell occupied by a waterdrop, the waterdrop's size increases by the number of the small drops in this cell, and these small drops disappears.
You are given a game and a position (
Each line of the following
The next line contains two integers
It is guaranteed that all the positions in the input are distinct.
Multiple test cases (about 100 cases), please read until EOF (End Of File).
If the
If the
4 4 5 10 2 1 4 2 3 3 2 4 4 3 1 2 4 3 4 4 4
0 5 0 3 0 2 1 3 0 1
#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
struct node{
bool exist;
int drops,time;
}map[111][111];
struct water{
int dir;
int wx,wy;
};
queue<water> q1;
queue<water> q2;
void broken(int x,int y){
map[x][y].exist=false;
water w;w.wx=x;w.wy=y;
w.dir=1;q1.push(w);
w.dir=2;q1.push(w);
w.dir=3;q1.push(w);
w.dir=4;q1.push(w);
}
bool panduan(water w,int k){
if(map[w.wx][w.wy].exist==true){
map[w.wx][w.wy].drops++;
map[w.wx][w.wy].time=k;
return true;
}
return false;
}
int x[111],y[111],val[111];
int r,c,n,t;
int main(int argc, char *argv[])
{
while(scanf("%d%d%d%d",&r,&c,&n,&t)!=EOF){
int i,j;
for(i=0;i<111;i++){
for(j=0;j<111;j++) {
map[i][j].drops=map[i][j].time=0;
map[i][j].exist=false;
}
}
for(i=0;i<n;i++){
scanf("%d%d%d",&x[i],&y[i],&val[i]);
map[x[i]][y[i]].exist=true;
map[x[i]][y[i]].drops=val[i];
}
int sx,sy;
scanf("%d%d",&sx,&sy);
int k;
broken(sx,sy);
for(k=1;k<=t;k++){
while(!q1.empty()){
water w=q1.front();
q1.pop();
if(w.dir==1){
if(w.wx>1) {w.wx--;q2.push(w);}
}
if(w.dir==2){
if(w.wx<r) {w.wx++;q2.push(w);}
}
if(w.dir==3){
if(w.wy>1) {w.wy--;q2.push(w);}
}
if(w.dir==4){
if(w.wy<c) {w.wy++;q2.push(w);}
}
}
while(!q2.empty()){
water w=q2.front();
q2.pop();
if(panduan(w,k)==false) q1.push(w);
}
for(i=0;i<n;i++){
if(map[x[i]][y[i]].exist==true&&map[x[i]][y[i]].drops>=5) broken(x[i],y[i]);
}
}
while(!q1.empty()) q1.pop();
while(!q2.empty()) q2.pop();
for(i=0;i<n;i++){
if(map[x[i]][y[i]].exist==false) printf("0 %d\n",map[x[i]][y[i]].time);
else printf("1 %d\n",map[x[i]][y[i]].drops);
}
}
return 0;
}