India and China Origins
Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 676 Accepted Submission(s): 227
long time ago there are no himalayas between India and China, the both
cultures are frequently exchanged and are kept in sync at that time, but
eventually himalayas rise up. With that at first the communation
started to reduce and eventually died.
Let's
assume from my crude drawing that the only way to reaching from India
to China or viceversa is through that grid, blue portion is the ocean
and people haven't yet invented the ship. and the yellow portion is
desert and has ghosts roaming around so people can't travel that way.
and the black portions are the location which have mountains and white
portions are plateau which are suitable for travelling. moutains are
very big to get to the top, height of these mountains is infinite. So if
there is mountain between two white portions you can't travel by
climbing the mountain.
And at each step people can go to 4 adjacent positions.
Our
archeologists have taken sample of each mountain and estimated at which
point they rise up at that place. So given the times at which each
mountains rised up you have to tell at which time the communication
between India and China got completely cut off.
For each test case, the first line contains two space seperated integers N,M. next N lines consists of strings composed of 0,1 characters. 1 denoting that there's already a mountain at that place, 0 denoting the plateau. on N+2 line there will be an integer Q denoting the number of mountains that rised up in the order of times. Next Q lines contain 2 space seperated integers X,Y denoting that at ith year a mountain rised up at location X,Y.
T≤10
1≤N≤500
1≤M≤500
1≤Q≤N∗M
0≤X<N
0≤Y<M
print -1 if these two countries still connected in the end.
Hint:
From the picture above, we can see that China and India have no communication since 4th year.
4 6
011010
000010
100001
001000
7
0 3
1 5
1 3
0 0
1 2
2 4
2 1
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn=;
int fa[maxn];
bool out[maxn];
int X[maxn],Y[maxn];
int map[][];
char s[];
int find(int x){
return fa[x]==x?x:fa[x]=find(fa[x]);
} int main(){
int T,R,C,m;
scanf("%d",&T);
while(T--){
scanf("%d%d",&R,&C);
for(int i=C+;i<=(R+)*C;i++)fa[i]=i;
for(int i=;i<=C;i++)fa[i]=;
for(int i=C*(R+)+;i<=C*(R+);i++)fa[i]=C*(R+)+; for(int i=;i<=R;i++){
scanf("%s",s+);
for(int j=;j<=C;j++)
map[i][j]=s[j]-'';
} scanf("%d",&m);
for(int i=;i<=m;i++){
scanf("%d%d",&X[i],&Y[i]);X[i]++;Y[i]++;
map[X[i]][Y[i]]=;
} for(int i=;i<=R;i++)
for(int j=;j<C;j++){
if(map[i][j])continue;
if(map[i-][j]==)fa[find((i-)*C+j)]=find(i*C+j);
if(map[i+][j]==)fa[find((i+)*C+j)]=find(i*C+j);
if(map[i][j+]==)fa[find(i*C+j+)]=find(i*C+j);
if(map[i][j-]==)fa[find(i*C+j-)]=find(i*C+j);
} int ans=;
for(int i=m;i>=;i--){
if(find()==find(C*(R+)+)){
ans=i+;
break;
}
map[X[i]][Y[i]]=;
if(map[X[i]-][Y[i]]==)fa[find((X[i]-)*C+Y[i])]=find(X[i]*C+Y[i]);
if(map[X[i]+][Y[i]]==)fa[find((X[i]+)*C+Y[i])]=find(X[i]*C+Y[i]);
if(map[X[i]][Y[i]+]==)fa[find(X[i]*C+Y[i]+)]=find(X[i]*C+Y[i]);
if(map[X[i]][Y[i]-]==)fa[find(X[i]*C+Y[i]-)]=find(X[i]*C+Y[i]);
}
if(ans==m+)
printf("-1\n");
else
printf("%d\n",ans);
}
return ;
}