Codeforces Round #439 (Div. 2) E. The Untended Antiquity(Hash)
2017-10-09 16:21
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题目链接:http://codeforces.com/problemset/problem/869/E
wa,因为没有矩形相交的情况,所以直接Hash一下就行了。对于一个矩阵,针对四个顶点进行Hash,就可以判断两个点是否在同一个矩阵内就可以了,直接用随机数Hash也没有撞。。。
代码:
#include<bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
using namespace std;
const int MAXN=2505;
typedef pair<pair<int,int>,pair<int,int> > piiii;
typedef pair<int,int> pii;
typedef long long ll;
struct BIT
{
ll d[MAXN][MAXN];
int lowbit(int x)
{
return x&(-x);
}
void update(int x,int y,int w)
{
for(int i=x;i<MAXN;i+=lowbit(i))
for(int j=y;j<MAXN;j+=lowbit(j))
d[i][j]+=w;
}
ll query(int x,int y)
{
long long ans=0;
for(int i=x;i;i-=lowbit(i))
for(int j=y;j;j-=lowbit(j))
ans+=d[i][j];
return ans;
}
}B;
ll getR()
{
return (((((1LL*rand()<<16)+rand())<<16)+rand())<<16)+rand();
}
map<piiii,ll> M;
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
srand(233);
int n,m,q;
scanf("%d%d%d",&n,&m,&q);
for(int i=1;i<=q;i++)
{
int t,r1,r2,c1,c2;
scanf("%d%d%d%d%d",&t,&r1,&c1,&r2,&c2);
if(t==1)
{
ll val=getR();
M[mp(mp(r1,c1),mp(r2,c2))]=val;
B.update(r1,c1,val);
B.update(r1,c2+1,-val);
B.update(r2+1,c1,-val);
B.update(r2+1,c2+1,val);
}
else if(t==2)
{
ll val=M[mp(mp(r1,c1),mp(r2,c2))];
B.update(r1,c1,-val);
B.update(r1,c2+1,val);
B.update(r2+1,c1,val);
B.update(r2+1,c2+1,-val);
}
else
{
if(B.query(r1,c1)==B.query(r2,c2))
puts("Yes");
else
puts("No");
}
}
return 0;
}
wa,因为没有矩形相交的情况,所以直接Hash一下就行了。对于一个矩阵,针对四个顶点进行Hash,就可以判断两个点是否在同一个矩阵内就可以了,直接用随机数Hash也没有撞。。。
代码:
#include<bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
using namespace std;
const int MAXN=2505;
typedef pair<pair<int,int>,pair<int,int> > piiii;
typedef pair<int,int> pii;
typedef long long ll;
struct BIT
{
ll d[MAXN][MAXN];
int lowbit(int x)
{
return x&(-x);
}
void update(int x,int y,int w)
{
for(int i=x;i<MAXN;i+=lowbit(i))
for(int j=y;j<MAXN;j+=lowbit(j))
d[i][j]+=w;
}
ll query(int x,int y)
{
long long ans=0;
for(int i=x;i;i-=lowbit(i))
for(int j=y;j;j-=lowbit(j))
ans+=d[i][j];
return ans;
}
}B;
ll getR()
{
return (((((1LL*rand()<<16)+rand())<<16)+rand())<<16)+rand();
}
map<piiii,ll> M;
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
srand(233);
int n,m,q;
scanf("%d%d%d",&n,&m,&q);
for(int i=1;i<=q;i++)
{
int t,r1,r2,c1,c2;
scanf("%d%d%d%d%d",&t,&r1,&c1,&r2,&c2);
if(t==1)
{
ll val=getR();
M[mp(mp(r1,c1),mp(r2,c2))]=val;
B.update(r1,c1,val);
B.update(r1,c2+1,-val);
B.update(r2+1,c1,-val);
B.update(r2+1,c2+1,val);
}
else if(t==2)
{
ll val=M[mp(mp(r1,c1),mp(r2,c2))];
B.update(r1,c1,-val);
B.update(r1,c2+1,val);
B.update(r2+1,c1,val);
B.update(r2+1,c2+1,-val);
}
else
{
if(B.query(r1,c1)==B.query(r2,c2))
puts("Yes");
else
puts("No");
}
}
return 0;
}
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