10.17T1 联通块
缺钱的LZY
Problem BackGround
Lzy是一名优秀的中学生，这天，他又开始肝Azur Line了。在手游中，如何能够变得更 强，唯有氪金，Lzy深谙此道。正巧，Azur Line国庆礼包上线了，Lzy的忍痛割手，决定使用他的 多年积蓄。
众所周知，Lzy是一个聪明屁事多的好孩子，他把他所有的存钱罐的钥匙放入了他的存钱罐中，他 必须打开所有存钱罐以获得足够的钱，每个存钱罐可以被相应的钥匙打开或者被砸开，
但是，Lzy太优秀了，他深知砸开存钱罐是对广大劳动人民的极大不敬，所以，他想，在打开所有 存钱罐时，砸开最少的存钱罐。
由于Lzy还要忙着去Azur Line肝3-4，这个简单的问题就交给你了。
ProblemDescription
Lzy有n个存钱罐，Lzy已经将所有钥匙放入了一些存钱罐中，现在Lzy知道他每个钥匙所 在的存钱罐，Lzy想知道他最少砸几个存钱罐才能打开所有存钱罐。
Input
第一行一个整数n
第二行n个整数，第i个整数ai表示Lzy第i个存钱罐的钥匙放在了第ai个存钱罐中
Ouput
一个整数表示Lzy最少砸几个存钱罐才能打开所有存钱罐
SampleInput
4
2
1
2
4
Sample Output
2
Data Scale
|
测试点 |
n |
|
1~2 |
≤20 |
|
3~6 |
≤5000 |
|
7~9 |
≤1000000 |
|
10 |
≤5000000 |
Hint
选手下发文件提供了FastIO
此题本身就个裸的求联通块，但是LZY卡读入，所以只能写FastIO
code：
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #define N 10000005 5 using namespace std; 6 namespace fastIO{ 7 #define BUF_SIZE 100000 8 #define OUT_SIZE 100000 9 #define ll long long 10 //fread->read 11 bool IOerror=0; 12 inline char nc(){ 13 static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE; 14 if (p1==pend){ 15 p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin); 16 if (pend==p1){IOerror=1;return -1;} 17 //{printf("IO error!\n");system("pause");for (;;);exit(0);} 18 } 19 return *p1++; 20 } 21 inline bool blank(char ch){return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';} 22 inline void read(int &x){ 23 bool sign=0; char ch=nc(); x=0; 24 for (;blank(ch);ch=nc()); 25 if (IOerror)return; 26 if (ch=='-')sign=1,ch=nc(); 27 for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0'; 28 if (sign)x=-x; 29 } 30 inline void read(ll &x){ 31 bool sign=0; char ch=nc(); x=0; 32 for (;blank(ch);ch=nc()); 33 if (IOerror)return; 34 if (ch=='-')sign=1,ch=nc(); 35 for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0'; 36 if (sign)x=-x; 37 } 38 inline void read(double &x){ 39 bool sign=0; char ch=nc(); x=0; 40 for (;blank(ch);ch=nc()); 41 if (IOerror)return; 42 if (ch=='-')sign=1,ch=nc(); 43 for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0'; 44 if (ch=='.'){ 45 double tmp=1; ch=nc(); 46 for (;ch>='0'&&ch<='9';ch=nc())tmp/=10.0,x+=tmp*(ch-'0'); 47 } 48 if (sign)x=-x; 49 } 50 inline void read(char *s){ 51 char ch=nc(); 52 for (;blank(ch);ch=nc()); 53 if (IOerror)return; 54 for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch; 55 *s=0; 56 } 57 inline void read(char &c){ 58 for (c=nc();blank(c);c=nc()); 59 if (IOerror){c=-1;return;} 60 } 61 //getchar->read 62 inline void read1(int &x){ 63 char ch;int bo=0;x=0; 64 for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1; 65 for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar()); 66 if (bo)x=-x; 67 } 68 inline void read1(ll &x){ 69 char ch;int bo=0;x=0; 70 for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1; 71 for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar()); 72 if (bo)x=-x; 73 } 74 inline void read1(double &x){ 75 char ch;int bo=0;x=0; 76 for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1; 77 for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar()); 78 if (ch=='.'){ 79 double tmp=1; 80 for (ch=getchar();ch>='0'&&ch<='9';tmp/=10.0,x+=tmp*(ch-'0'),ch=getchar()); 81 } 82 if (bo)x=-x; 83 } 84 inline void read1(char *s){ 85 char ch=getchar(); 86 for (;blank(ch);ch=getchar()); 87 for (;!blank(ch);ch=getchar())*s++=ch; 88 *s=0; 89 } 90 inline void read1(char &c){for (c=getchar();blank(c);c=getchar());} 91 //scanf->read 92 inline void read2(int &x){scanf("%d",&x);} 93 inline void read2(ll &x){ 94 #ifdef _WIN32 95 scanf("%I64d",&x); 96 #else 97 #ifdef __linux 98 scanf("%lld",&x); 99 #else 100 puts("error:can't recognize the system!"); 101 #endif 102 #endif 103 } 104 inline void read2(double &x){scanf("%lf",&x);} 105 inline void read2(char *s){scanf("%s",s);} 106 inline void read2(char &c){scanf(" %c",&c);} 107 inline void readln2(char *s){gets(s);} 108 //fwrite->write 109 struct Ostream_fwrite{ 110 char *buf,*p1,*pend; 111 Ostream_fwrite(){buf=new char[BUF_SIZE];p1=buf;pend=buf+BUF_SIZE;} 112 void out(char ch){ 113 if (p1==pend){ 114 fwrite(buf,1,BUF_SIZE,stdout);p1=buf; 115 } 116 *p1++=ch; 117 } 118 void print(int x){ 119 static char s[15],*s1;s1=s; 120 if (!x)*s1++='0';if (x<0)out('-'),x=-x; 121 while(x)*s1++=x%10+'0',x/=10; 122 while(s1--!=s)out(*s1); 123 } 124 void println(int x){ 125 static char s[15],*s1;s1=s; 126 if (!x)*s1++='0';if (x<0)out('-'),x=-x; 127 while(x)*s1++=x%10+'0',x/=10; 128 while(s1--!=s)out(*s1); out('\n'); 129 } 130 void print(ll x){ 131 static char s[25],*s1;s1=s; 132 if (!x)*s1++='0';if (x<0)out('-'),x=-x; 133 while(x)*s1++=x%10+'0',x/=10; 134 while(s1--!=s)out(*s1); 135 } 136 void println(ll x){ 137 static char s[25],*s1;s1=s; 138 if (!x)*s1++='0';if (x<0)out('-'),x=-x; 139 while(x)*s1++=x%10+'0',x/=10; 140 while(s1--!=s)out(*s1); out('\n'); 141 } 142 void print(double x,int y){ 143 static ll mul[]={1,10,100,1000,10000,100000,1000000,10000000,100000000, 144 1000000000,10000000000LL,100000000000LL,1000000000000LL,10000000000000LL, 145 100000000000000LL,1000000000000000LL,10000000000000000LL,100000000000000000LL}; 146 if (x<-1e-12)out('-'),x=-x;x*=mul[y]; 147 ll x1=(ll)floor(x); if (x-floor(x)>=0.5)++x1; 148 ll x2=x1/mul[y],x3=x1-x2*mul[y]; print(x2); 149 if (y>0){out('.'); for (size_t i=1;i<y&&x3*mul[i]<mul[y];out('0'),++i); print(x3);} 150 } 151 void println(double x,int y){print(x,y);out('\n');} 152 void print(char *s){while (*s)out(*s++);} 153 void println(char *s){while (*s)out(*s++);out('\n');} 154 void flush(){if (p1!=buf){fwrite(buf,1,p1-buf,stdout);p1=buf;}} 155 ~Ostream_fwrite(){flush();} 156 }Ostream; 157 inline void print(int x){Ostream.print(x);} 158 inline void println(int x){Ostream.println(x);} 159 inline void print(char x){Ostream.out(x);} 160 inline void println(char x){Ostream.out(x);Ostream.out('\n');} 161 inline void print(ll x){Ostream.print(x);} 162 inline void println(ll x){Ostream.println(x);} 163 inline void print(double x,int y){Ostream.print(x,y);} 164 inline void println(double x,int y){Ostream.println(x,y);} 165 inline void print(char *s){Ostream.print(s);} 166 inline void println(char *s){Ostream.println(s);} 167 inline void println(){Ostream.out('\n');} 168 inline void flush(){Ostream.flush();} 169 //puts->write 170 char Out[OUT_SIZE],*o=Out; 171 inline void print1(int x){ 172 static char buf[15]; 173 char *p1=buf;if (!x)*p1++='0';if (x<0)*o++='-',x=-x; 174 while(x)*p1++=x%10+'0',x/=10; 175 while(p1--!=buf)*o++=*p1; 176 } 177 inline void println1(int x){print1(x);*o++='\n';} 178 inline void print1(ll x){ 179 static char buf[25]; 180 char *p1=buf;if (!x)*p1++='0';if (x<0)*o++='-',x=-x; 181 while(x)*p1++=x%10+'0',x/=10; 182 while(p1--!=buf)*o++=*p1; 183 } 184 inline void println1(ll x){print1(x);*o++='\n';} 185 inline void print1(char c){*o++=c;} 186 inline void println1(char c){*o++=c;*o++='\n';} 187 inline void print1(char *s){while (*s)*o++=*s++;} 188 inline void println1(char *s){print1(s);*o++='\n';} 189 inline void println1(){*o++='\n';} 190 inline void flush1(){if (o!=Out){if (*(o-1)=='\n')*--o=0;puts(Out);}} 191 struct puts_write{ 192 ~puts_write(){flush1();} 193 }_puts; 194 inline void print2(int x){printf("%d",x);} 195 inline void println2(int x){printf("%d\n",x);} 196 inline void print2(char x){printf("%c",x);} 197 inline void println2(char x){printf("%c\n",x);} 198 inline void print2(ll x){ 199 #ifdef _WIN32 200 printf("%I64d",x); 201 #else 202 #ifdef __linux 203 printf("%lld",x); 204 #else 205 puts("error:can't recognize the system!"); 206 #endif 207 #endif 208 } 209 inline void println2(ll x){print2(x);printf("\n");} 210 inline void println2(){printf("\n");} 211 #undef ll 212 #undef OUT_SIZE 213 #undef BUF_SIZE 214 }; 215 using namespace fastIO; 216 struct node{ 217 int u,v; 218 }e[N]; 219 int vis[N],first[N],nxt[N],cnt; 220 inline void add(int u,int v){ 221 e[++cnt].u=u; 222 e[cnt].v=v; 223 nxt[cnt]=first[u]; 224 first[u]=cnt; 225 } 226 inline void dfs(int x){ 227 vis[x]=1; 228 for(int i=first[x];i;i=nxt[i]){ 229 int v=e[i].v; 230 if(vis[v])continue; 231 dfs(v); 232 } 233 } 234 int main(){ 235 // freopen("lzy.in","r",stdin); 236 // freopen("lzy.out","w",stdout); 237 int n; 238 read(n); 239 for(int i=1;i<=n;++i){ 240 int x; 241 read(x); 242 add(i,x); 243 add(x,i); 244 } 245 int ans=0; 246 for(int i=1;i<=n;++i){ 247 if(vis[i])continue; 248 ans++; 249 dfs(i); 250 } 251 cout<<ans; 252 return 0; 253 }
over
