• 1810 连续区间(分治)


    基准时间限制:1.5 秒 空间限制:131072 KB 分值: 80 难度:5级算法题
    区间内所有元素排序后,任意相邻两个元素值差为1的区间称为“连续区间”
    如:3,1,2是连续区间,但3,1,4不是连续区间
    给出一个1~n的排列,求出有多少个连续区间
    Input
    一个数n(n<=1,000,000)
    第二行n个数,表示一个1~n的排列
    Output
    一个数,表示有多少个连续区间
    Input示例
    5
    2 1 5 3 4
    Output示例
    9
    样例解释:
    区间[1,1][2,2][3,3][4,4][5,5][1,2][4,5][3,4][1,5]为连续区间//[l,r]表示从第l个数到第r个数构成的区间


    // 容易想到分治,但是还是没想出怎么细节,唉


    写得很详细了,比较好理解,但是,写起来还是要注意挺多地方的,细节很多,写了好久,唉
    大部分是超快读入挂
      1 #include <bits/stdc++.h>
      2 using namespace std;
      3 #define MOD 1000000007
      4 #define INF 0x3f3f3f3f
      5 #define eps 1e-9
      6 #define LL long long
      7 #define MX 1000005
      8 
      9 namespace fastIO{
     10     #define BUF_SIZE 100000
     11     #define OUT_SIZE 100000
     12     #define ll long long
     13     //fread->read
     14     bool IOerror=0;
     15     inline char nc(){
     16         static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
     17         if (p1==pend){
     18             p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin);
     19             if (pend==p1){IOerror=1;return -1;}
     20             //{printf("IO error!
    ");system("pause");for (;;);exit(0);}
     21         }
     22         return *p1++;
     23     }
     24     inline bool blank(char ch){return ch==' '||ch=='
    '||ch=='
    '||ch=='	';}
     25     inline void read(int &x){
     26         bool sign=0; char ch=nc(); x=0;
     27         for (;blank(ch);ch=nc());
     28         if (IOerror)return;
     29         if (ch=='-')sign=1,ch=nc();
     30         for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
     31         if (sign)x=-x;
     32     }
     33     inline void read(ll &x){
     34         bool sign=0; char ch=nc(); x=0;
     35         for (;blank(ch);ch=nc());
     36         if (IOerror)return;
     37         if (ch=='-')sign=1,ch=nc();
     38         for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
     39         if (sign)x=-x;
     40     }
     41     inline void read(double &x){
     42         bool sign=0; char ch=nc(); x=0;
     43         for (;blank(ch);ch=nc());
     44         if (IOerror)return;
     45         if (ch=='-')sign=1,ch=nc();
     46         for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
     47         if (ch=='.'){
     48             double tmp=1; ch=nc();
     49             for (;ch>='0'&&ch<='9';ch=nc())tmp/=10.0,x+=tmp*(ch-'0');
     50         }
     51         if (sign)x=-x;
     52     }
     53     inline void read(char *s){
     54         char ch=nc();
     55         for (;blank(ch);ch=nc());
     56         if (IOerror)return;
     57         for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch;
     58         *s=0;
     59     }
     60     inline void read(char &c){
     61         for (c=nc();blank(c);c=nc());
     62         if (IOerror){c=-1;return;}
     63     }
     64     //getchar->read
     65     inline void read1(int &x){
     66         char ch;int bo=0;x=0;
     67         for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;
     68         for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
     69         if (bo)x=-x;
     70     }
     71     inline void read1(ll &x){
     72         char ch;int bo=0;x=0;
     73         for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;
     74         for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
     75         if (bo)x=-x;
     76     }
     77     inline void read1(double &x){
     78         char ch;int bo=0;x=0;
     79         for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;
     80         for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
     81         if (ch=='.'){
     82             double tmp=1;
     83             for (ch=getchar();ch>='0'&&ch<='9';tmp/=10.0,x+=tmp*(ch-'0'),ch=getchar());
     84         }
     85         if (bo)x=-x;
     86     }
     87     inline void read1(char *s){
     88         char ch=getchar();
     89         for (;blank(ch);ch=getchar());
     90         for (;!blank(ch);ch=getchar())*s++=ch;
     91         *s=0;
     92     }
     93     inline void read1(char &c){for (c=getchar();blank(c);c=getchar());}
     94     //scanf->read
     95     inline void read2(int &x){scanf("%d",&x);}
     96     inline void read2(ll &x){
     97         #ifdef _WIN32
     98             scanf("%I64d",&x);
     99         #else
    100         #ifdef __linux
    101             scanf("%lld",&x);
    102         #else
    103             puts("error:can't recognize the system!");
    104         #endif
    105         #endif
    106     }
    107     inline void read2(double &x){scanf("%lf",&x);}
    108     inline void read2(char *s){scanf("%s",s);}
    109     inline void read2(char &c){scanf(" %c",&c);}
    110     inline void readln2(char *s){gets(s);}
    111     //fwrite->write
    112     struct Ostream_fwrite{
    113         char *buf,*p1,*pend;
    114         Ostream_fwrite(){buf=new char[BUF_SIZE];p1=buf;pend=buf+BUF_SIZE;}
    115         void out(char ch){
    116             if (p1==pend){
    117                 fwrite(buf,1,BUF_SIZE,stdout);p1=buf;
    118             }
    119             *p1++=ch;
    120         }
    121         void print(int x){
    122             static char s[15],*s1;s1=s;
    123             if (!x)*s1++='0';if (x<0)out('-'),x=-x;
    124             while(x)*s1++=x%10+'0',x/=10;
    125             while(s1--!=s)out(*s1);
    126         }
    127         void println(int x){
    128             static char s[15],*s1;s1=s;
    129             if (!x)*s1++='0';if (x<0)out('-'),x=-x;
    130             while(x)*s1++=x%10+'0',x/=10;
    131             while(s1--!=s)out(*s1); out('
    ');
    132         }
    133         void print(ll x){
    134             static char s[25],*s1;s1=s;
    135             if (!x)*s1++='0';if (x<0)out('-'),x=-x;
    136             while(x)*s1++=x%10+'0',x/=10;
    137             while(s1--!=s)out(*s1);
    138         }
    139         void println(ll x){
    140             static char s[25],*s1;s1=s;
    141             if (!x)*s1++='0';if (x<0)out('-'),x=-x;
    142             while(x)*s1++=x%10+'0',x/=10;
    143             while(s1--!=s)out(*s1); out('
    ');
    144         }
    145         void print(double x,int y){
    146             static ll mul[]={1,10,100,1000,10000,100000,1000000,10000000,100000000,
    147                 1000000000,10000000000LL,100000000000LL,1000000000000LL,10000000000000LL,
    148                 100000000000000LL,1000000000000000LL,10000000000000000LL,100000000000000000LL};
    149             if (x<-1e-12)out('-'),x=-x;x*=mul[y];
    150             ll x1=(ll)floor(x); if (x-floor(x)>=0.5)++x1;
    151             ll x2=x1/mul[y],x3=x1-x2*mul[y]; print(x2);
    152             if (y>0){out('.'); for (size_t i=1;i<y&&x3*mul[i]<mul[y];out('0'),++i); print(x3);}
    153         }
    154         void println(double x,int y){print(x,y);out('
    ');}
    155         void print(char *s){while (*s)out(*s++);}
    156         void println(char *s){while (*s)out(*s++);out('
    ');}
    157         void flush(){if (p1!=buf){fwrite(buf,1,p1-buf,stdout);p1=buf;}}
    158         ~Ostream_fwrite(){flush();}
    159     }Ostream;
    160     inline void print(int x){Ostream.print(x);}
    161     inline void println(int x){Ostream.println(x);}
    162     inline void print(char x){Ostream.out(x);}
    163     inline void println(char x){Ostream.out(x);Ostream.out('
    ');}
    164     inline void print(ll x){Ostream.print(x);}
    165     inline void println(ll x){Ostream.println(x);}
    166     inline void print(double x,int y){Ostream.print(x,y);}
    167     inline void println(double x,int y){Ostream.println(x,y);}
    168     inline void print(char *s){Ostream.print(s);}
    169     inline void println(char *s){Ostream.println(s);}
    170     inline void println(){Ostream.out('
    ');}
    171     inline void flush(){Ostream.flush();}
    172     //puts->write
    173     char Out[OUT_SIZE],*o=Out;
    174     inline void print1(int x){
    175         static char buf[15];
    176         char *p1=buf;if (!x)*p1++='0';if (x<0)*o++='-',x=-x;
    177         while(x)*p1++=x%10+'0',x/=10;
    178         while(p1--!=buf)*o++=*p1;
    179     }
    180     inline void println1(int x){print1(x);*o++='
    ';}
    181     inline void print1(ll x){
    182         static char buf[25];
    183         char *p1=buf;if (!x)*p1++='0';if (x<0)*o++='-',x=-x;
    184         while(x)*p1++=x%10+'0',x/=10;
    185         while(p1--!=buf)*o++=*p1;
    186     }
    187     inline void println1(ll x){print1(x);*o++='
    ';}
    188     inline void print1(char c){*o++=c;}
    189     inline void println1(char c){*o++=c;*o++='
    ';}
    190     inline void print1(char *s){while (*s)*o++=*s++;}
    191     inline void println1(char *s){print1(s);*o++='
    ';}
    192     inline void println1(){*o++='
    ';}
    193     inline void flush1(){if (o!=Out){if (*(o-1)=='
    ')*--o=0;puts(Out);}}
    194     struct puts_write{
    195         ~puts_write(){flush1();}
    196     }_puts;
    197     inline void print2(int x){printf("%d",x);}
    198     inline void println2(int x){printf("%d
    ",x);}
    199     inline void print2(char x){printf("%c",x);}
    200     inline void println2(char x){printf("%c
    ",x);}
    201     inline void print2(ll x){
    202         #ifdef _WIN32
    203             printf("%I64d",x);
    204         #else
    205         #ifdef __linux
    206             printf("%lld",x);
    207         #else
    208             puts("error:can't recognize the system!");
    209         #endif
    210         #endif
    211     }
    212     inline void println2(ll x){print2(x);printf("
    ");}
    213     inline void println2(){printf("
    ");}
    214     #undef ll
    215     #undef OUT_SIZE
    216     #undef BUF_SIZE
    217 };
    218 using namespace fastIO;
    219 
    220 int n;
    221 int dat[MX];
    222 int tt[MX*2];
    223 int ma[MX],mi[MX];
    224 LL ans;
    225 
    226 void func(int l,int r)
    227 {
    228     int mid = (l+r)/2;
    229     ma[mid] = mi[mid] = dat[mid];
    230     for (int i=mid-1;i>=l;i--)
    231     {
    232         ma[i] = max(ma[i+1],dat[i]);
    233         mi[i] = min(mi[i+1],dat[i]);
    234     }
    235     ma[mid+1] = mi[mid+1] = dat[mid+1];
    236     for (int i=mid+2;i<=r;i++)
    237     {
    238         ma[i] = max(ma[i-1],dat[i]);
    239         mi[i] = min(mi[i-1],dat[i]);
    240     }
    241     for (int i=l;i<=mid;i++)    // 都在左边
    242     {
    243         int j = ma[i] - mi[i] + i;
    244         if (j>mid&&j<=r&&ma[j]<ma[i]&&mi[j]>mi[i]) ans++;
    245     }
    246     for (int j=mid+1;j<=r;j++)  // 都在右边
    247     {
    248         int i = -(ma[j] - mi[j]) + j;
    249         if (i>=l&&i<=mid&&ma[i]<ma[j]&&mi[i]>mi[j]) ans++;
    250     }
    251     //左大 右小
    252     int l_=mid+1, r_=mid;
    253     for (int i=mid;i>=l;i--)
    254     {
    255         while (r_<r&&ma[r_+1]<ma[i])
    256         {
    257             r_++;
    258             tt[mi[r_]+r_]++;
    259         }
    260         while(l_<=r_&&mi[l_]>mi[i])
    261         {
    262             tt[mi[l_]+l_]--;
    263             l_++;
    264         }
    265         ans += tt[ma[i] + i];
    266     }
    267     while(l_<=r_) {tt[mi[l_]+l_]--;l_++;}
    268     //左小,右大
    269     l_ = mid+1, r_=mid;
    270     for (int j=mid+1;j<=r;j++)
    271     {
    272         while (l_>l&&ma[l_-1]<ma[j])
    273         {
    274             l_--;
    275             tt[mi[l_]-l_+MX]++;
    276         }
    277         while(r_>=l_&&mi[r_]>mi[j])
    278         {
    279             tt[mi[r_]-r_+MX]--;
    280             r_--;
    281         }
    282         ans += tt[ma[j]-j+MX];
    283     }
    284     while(r_>=l_) {tt[mi[r_]-r_+MX]--; r_--;}
    285 
    286     if (l==r) return ;
    287     func(l,mid); func(mid+1,r);
    288 }
    289 
    290 int main()
    291 {
    292     fastIO::read(n);
    293     for (int i=1;i<=n;i++)
    294         fastIO::read(dat[i]);
    295     func(1,n);
    296     printf("%lld
    ",ans+n);
    297     return 0;
    298 }
    View Code
    
    
    



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  • 原文地址:https://www.cnblogs.com/haoabcd2010/p/7635092.html
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