Wireless Network
题意
N台损坏的计算机，任意两台计算机之间距离小于D即可连接，计算机最初不连接，经过若干次操作，操作一，O X ，修复X计算机，并连接所有与X距离不大于D的计算机，操作二，S P Q ，询问P，Q是否存在一条连接的线路，不存在输出FAIL，存在输出SUCCESS
题解
并查集求解
vis ：记录是否修复
ms ：记录任意两点距离
per ：记录祖先节点
#pragma GCC optimize(2)
#include <iostream>
#include <map>
#include <vector>
#include <queue>
#include <string>
#include <set>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <list>
#include <deque>
#include <queue>
#include <stack>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <iomanip>
using namespace std;
const int inf=0x3f3f3f3f;
#define eps 1e-5
#define pii pair<int,int>
#define FI first
#define SE second
#define ll long long
#define ull unsigned long long
const ll mod = 4294967296;/// 998244353;
const int mxn = 1e3 +7;
int _ , n , m , t , k , cnt , si , res ,ans ;
template <class T>
void rd(T &x){
T flag = 1 ; x = 0 ; char ch = getchar() ;
while(!isdigit(ch)) { if(ch=='-') flag = -1; ch = getchar(); }
while(isdigit(ch)) { x = (x<<1) + (x<<3) + (ch^48); ch = getchar(); }
x*=flag;
}
int to[mxn] , nx[mxn] , head[mxn] ,per[mxn],x[mxn],y[mxn] ,ms[mxn][mxn];
char op;
bool vis[mxn] ;
int Find(int x) { return x==per[x]?x:per[x] = Find(per[x]); }
void Union(int u,int v) { per[ Find(u) ] = Find(v); }
int Dis(int i,int j){return (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]) ;}
void solve()
{
cin>>n>>k;
memset(vis,0,sizeof vis);
for(int i=1;i<=n;i++) cin>>x[i]>>y[i] , per[i] = i ;
for(int i=1;i<=n;i++) for(int j=i+1;j<=n;j++) ms[i][j] = ms[j][i] = (Dis(i,j)<=k*k?1:0);
while(cin>>op){
if(op=='O') {
cin>>si; vis[si] = 1 ;
for(int i=1;i<=n;i++) {
if(ms[si][i] && i!=si && vis[i]){
int ui = Find(si) , vi = Find(i);
if(ui!=vi) per[ui] = vi ;
}
}
} else {
int u,v,ui,vi;
cin>>u>>v;
ui = Find(u) , vi = Find(v) ;
if(ui==vi) cout<<"SUCCESS
";
else cout<<"FAIL
";
}
}
}
int main(){ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); solve();return 0;}
The Suspects
题意
求M个集合中，和 0号 在同一集合中所有不同编号的人的个数
题解
并查集
#pragma GCC optimize(2)
#include <iostream>
#include <map>
#include <vector>
#include <queue>
#include <string>
#include <set>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <list>
#include <deque>
#include <queue>
#include <stack>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <iomanip>
using namespace std;
const int inf=0x3f3f3f3f;
#define eps 1e-5
#define pii pair<int,int>
#define FI first
#define SE second
#define ll long long
#define ull unsigned long long
const ll mod = 4294967296;/// 998244353;
const int mxn = 3e4 +7;
int _ , n , m , t , k , cnt , si , res ,ans ;
template <class T>
void rd(T &x){
T flag = 1 ; x = 0 ; char ch = getchar() ;
while(!isdigit(ch)) { if(ch=='-') flag = -1; ch = getchar(); }
while(isdigit(ch)) { x = (x<<1) + (x<<3) + (ch^48); ch = getchar(); }
x*=flag;
}
int to[mxn] , nx[mxn] , head[mxn] ,per[mxn],x[mxn],y[mxn];
char op;
bool vis[mxn] ;
int Find(int x) { return x==per[x]?x:per[x] = Find(per[x]); }
void Union(int u,int v) { per[ Find(u) ] = Find(v); }
int Dis(int i,int j){return (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]) ;}
void solve()
{
while(cin>>n>>m&&(m||n)){
for(int i=0;i<=n;i++) per[i] = i ;
int u , v , ui , vi ; ans = 1 ;
for(int i=1;i<=m;i++){
for(cin>>k , _=1;k;k--){
cin>>v;
if(_==1) { u = v ; _=0 ; ui = Find(u);}
else {
vi = Find(v) ;
if(ui!=vi) per[vi] = ui ;
}
}
}
ui = Find(0) ;
for(int i=1;i<=n;i++) ans+= (Find(i)==ui);
cout<<ans<<endl;
}
}
int main(){ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); solve();return 0;}
How Many Tables
题意
求联通块的个数
题解
并查集暴搜
#pragma GCC optimize(2)
#include <iostream>
#include <map>
#include <vector>
#include <queue>
#include <string>
#include <set>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <list>
#include <deque>
#include <queue>
#include <stack>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <iomanip>
using namespace std;
const int inf=0x3f3f3f3f;
#define eps 1e-5
#define pii pair<int,int>
#define FI first
#define SE second
#define ll long long
#define ull unsigned long long
const ll mod = 4294967296;/// 998244353;
const int mxn = 3e4 +7;
int _ , n , m , t , k , cnt , si , res ,ans ;
template <class T>
void rd(T &x){
T flag = 1 ; x = 0 ; char ch = getchar() ;
while(!isdigit(ch)) { if(ch=='-') flag = -1; ch = getchar(); }
while(isdigit(ch)) { x = (x<<1) + (x<<3) + (ch^48); ch = getchar(); }
x*=flag;
}
int to[mxn] , nx[mxn] , head[mxn] ,per[mxn],x[mxn],y[mxn];
char op;
bool vis[mxn] ;
int Find(int x) { return x==per[x]?x:per[x] = Find(per[x]); }
void Union(int u,int v) { per[ Find(u) ] = Find(v); }
int Dis(int i,int j){return (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]) ;}
void solve()
{
for(cin>>t;t;t--){
cin>>n>>m; ans = 0 ;
for(int i=1;i<=n;i++) per[i] = i ;
for(int i=1,u,v;i<=m;i++){
cin>>u>>v;
int ui = Find(u) , vi = Find(v) ;
if(ui!=vi) per[ui] = vi ;
}
for(int i=1;i<=n;i++) if(per[i]==i) ans++;
cout<<ans<<endl;
}
}
int main(){ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); solve();return 0;}
食物链
题解
S:X相对于祖先节点Y的状态，即 0 ， 1 ， 2 ，表示方法为s[ X,Y ] = root
per：存储祖先节点
规定 S[T] = 0 :XY同类
S[T] = 1 :X捕食Y
S[T] = 2 :Y捕食X
初始化，per[x] = x ， 那么相对于祖先节点的关系为S[X] = 0
对于样例来说
2 1 2 ：S[1,1] = 0 , S[2,2] = 0 , S[ 1,2 ] = 2 , 而 per[1] = 1 , per[2] = 2
那么，per[2] = 1 , S[2,1] = S[1,2] = 1 .
2 3 3 ：S[2,3] = 2 , S[3,per[3] = 3] = s[3,3] = 0 , s[2,per[2]] = s[2,1] = 1 ，而per[2] = 2 , per[3] = 3
那么，per[3] = 1 , S[3,per[3]] = S[3,1] = S[2,3] - 1 + S[2,1] = 2 - 1 + 1 = 2
#pragma GCC optimize(2)
#include <iostream>
#include <map>
#include <vector>
#include <queue>
#include <string>
#include <set>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <list>
#include <deque>
#include <queue>
#include <stack>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <iomanip>
using namespace std;
const int inf=0x3f3f3f3f;
#define eps 1e-5
#define pii pair<int,int>
#define FI first
#define SE second
#define ll long long
#define ull unsigned long long
const ll mod = 4294967296;/// 998244353;
const int mxn = 3e5 +7;
int _ , n , m , t , k , cnt , si , res ,ans ;
template <class T>
void rd(T &x){
T flag = 1 ; x = 0 ; char ch = getchar() ;
while(!isdigit(ch)) { if(ch=='-') flag = -1; ch = getchar(); }
while(isdigit(ch)) { x = (x<<1) + (x<<3) + (ch^48); ch = getchar(); }
x*=flag;
}
int per[mxn] , sta[mxn];
int Find(int x)
{
if(x==per[x]) return x;
int top = Find(per[x]) ; sta[x] = ( (sta[ per[x] ]+sta[x])%3); per[x] = top ;
return top ;
}
void solve()
{
scanf("%d %d",&n,&m);
for(int i=0;i<=n;i++) per[i] = i , sta[i] = 0 ; ans = 0 ;
for(int i=1,u,v,op;i<=m;i++){
scanf("%d %d %d",&op,&u,&v);
if(u>n || v>n || (op==2 && u==v)) { ++ans;continue; }
int ui = Find(u) , vi = Find(v) ;
if(ui==vi && ( (op==1 && sta[u]!=sta[v]) || op==2 && (sta[u]+1)%3!=sta[v])) ++ans;
else per[vi] = ui , sta[vi] = ( -sta[v] + sta[u] + op + 3 - 1)%3;
}
printf("%d
",ans);
}
int main(){ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); solve();return 0;}
Parity game
#pragma GCC optimize(2)
#include <iostream>
#include <map>
#include <vector>
#include <queue>
#include <string>
#include <set>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <list>
#include <deque>
#include <queue>
#include <stack>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <iomanip>
using namespace std;
const int inf=0x3f3f3f3f;
#define eps 1e-5
#define pii pair<int,int>
#define FI first
#define SE second
#define ll long long
#define ull unsigned long long
const ll mod = 4294967296;/// 998244353;
const int mxn = 1e5 +7;
int _ , n , m , t , k , cnt , si , res ,ans ;
template <class T>
void rd(T &x){
T flag = 1 ; x = 0 ; char ch = getchar() ;
while(!isdigit(ch)) { if(ch=='-') flag = -1; ch = getchar(); }
while(isdigit(ch)) { x = (x<<1) + (x<<3) + (ch^48); ch = getchar(); }
x*=flag;
}
int per[mxn] , sta[mxn] , uni[mxn] ;
pair< pii,string > no[mxn];
int Find(int x){
if(x==per[x]) return per[x];
int top = Find(per[x]);
sta[x]^=sta[per[x]] ;
return per[x] = top;
}
void solve()
{
int ans = 0 , cnt = 0 ;
string str ;
cin>>n>>m;
for(int i=1;i<=m;i++) {
cin>>no[i].FI.FI>>no[i].FI.SE>>no[i].SE;
no[i].FI.FI--;
uni[++cnt] = no[i].FI.FI , uni[++cnt] = no[i].FI.SE;
}
sort(uni+1,uni+1+cnt);
int lim = unique(uni+1,uni+1+cnt) - uni - 1 ;
for(int i=1;i<=lim;i++) per[i] = i , sta[i] = 0 ;
for(int i=1;i<=m;i++){
int u = lower_bound( uni+1,uni+1+lim , no[i].FI.FI ) - uni ;
int v = lower_bound( uni+1,uni+1+lim , no[i].FI.SE ) - uni ;
int ui = Find(u) , vi = Find(v);
int bit = no[i].SE[0]=='o'?1 : 0 ;
if(ui==vi && sta[u]^bit!=sta[v]) break;
if(ui<vi) per[ui] = vi , sta[ui] = sta[v]^bit^sta[u];
else per[vi] = ui , sta[vi] = sta[u]^bit^sta[v];
++ans;
}
cout<<ans<<endl;
}
int main(){ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); solve();return 0;}
暴毙一上午的代码，至此不知道Find替换为下面的代码为什么会WA
int Find(int x){
if(x==per[x]) return per[x];
sta[x]^=sta[per[x]] ;
return per[x] = Find(per[x]);
}