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  • 二分刷题单

    下面的题目都还是挺不错的,因为比较简单,所以说就不写题解了,有疑问的可以去看题目的题解,这里只放代码:

    一道最长不下降子序列的题目(二分优化)

    /*
        Name:【P1233】木棍加工 
        Copyright: njc 
        Author: Mudrobot
        Date: 2018/10/12 15:29:47
        Description: greedy! 
    */
    #include<bits/stdc++.h>
    #define gc() getchar()//caution!!!
    #define N 20000
    using namespace std;
    /*inline char gc() {
      static char buf[1<<18],*fs,*ft;
      return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<18,stdin)),fs==ft)?EOF:*fs++;
    }*/
    template<class T>
    inline void read(T &aa) {
      register int k=0,f=1;
      register char c=gc();
      for (;!isdigit(c);c=gc()) if(c=='-')f=-1;
      for (;isdigit(c);c=gc()) k=(k<<3)+(k<<1)+(c-'0');
      aa=k*f;
    }
    template<class T>
    inline void out(T x){if(x>9)out(x/10);putchar(x%10+'0');}
    struct sd{int x,y;}stk[N];
    int n,mp[N];
    bool cmp(sd a,sd b){if(a.x>b.x) return true;if(a.x==b.x&&a.y>b.y)return true;return false;}
    
    int main()
    {
        read(n);
        for(int i=1;i<=n;++i) read(stk[i].x),read(stk[i].y);//scanf("%d%d",&stk[i].x,&stk[i].y);
        sort(stk+1,stk+1+n,cmp); int ans=0;
        for(int i=1;i<=n;++i)
        {
            if(stk[i].y>mp[ans])mp[++ans]=stk[i].y;
            else
            {
                int l=1,r=ans;
                while(l<=r)
                {
                    int mid=(l+r)/2;
                    if(mp[mid]>=stk[i].y)r=mid-1;
                    else l=mid+1;
                }
                mp[l]=stk[i].y;
            }
        }
        printf("%d",ans);
        return 0;
    }
    /*
    
    */
    
    

    二分好题,注意取斜率小于0的最右边和斜率大于0的左边进行二分!(差为0时就是我们要找的答案点)

    /*
        Name: P1663 山
        Copyright: njc
        Author: Mudrobot
        Date: 2018/10/20 15:39:54
        Description: Binary Search
    */
    #include<bits/stdc++.h>
    #define gc() getchar()//caution!!!
    #define N 5005
    #define INF 1000004
    #define eps 0.01
    using namespace std;
    /*inline char gc() {
      static char buf[1<<18],*fs,*ft;
      return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<18,stdin)),fs==ft)?EOF:*fs++;
    }*/
    template<class T>
    inline void read(T &aa) {
      register int k=0,f=1;
      register char c=gc();
      for (;!isdigit(c);c=gc()) if(c=='-')f=-1;
      for (;isdigit(c);c=gc()) k=(k<<3)+(k<<1)+(c-'0');
      aa=k*f;
    }
    template<class T>
    inline void out(T x){if(x>9)out(x/10);putchar(x%10+'0');}
    struct sd{
        double k,b;
    }line[N];
    int n,x[N],y[N];
    void construct(int a1,int a2){
        double x1=(double)x[a1],y1=(double)y[a1],x2=(double)x[a2],y2=(double)y[a2];
        line[a1].k=(y1-y2)/(x1-x2);line[a1].b=y1-line[a1].k*x1;
    }
    bool check(double y){
        bool flag1=false,flag2=false;
        double right=0,left=0;
        for(int i=1;i<n;++i){
            if(fabs(line[i].k)==0&&line[i].b>y) return false;
            if(line[i].k<0){
                if(!flag1){flag1=true;right=(y-line[i].b)/line[i].k;}
                else right=max(right,(y-line[i].b)/line[i].k);
            }
            else if(line[i].k>0){
                if(!flag2){flag2=true;left=((y-line[i].b)/line[i].k);}
                else left=min(left,(y-line[i].b)/line[i].k);
            }
        }
        if(left-right>=0) return true;
        return false;
    }
    int main()
    {
        //freopen(".in", "r", stdin);freopen(".out", "w", stdout);
        read(n);
        for(int i=1;i<=n;++i) read(x[i]),read(y[i]);
        for(int i=2;i<=n;++i) construct(i-1,i);
        //for(int i=1;i<n;++i) printf("%lf %lf
    ",line[i].k,line[i].b);
        double l=0,r=INF,final;
        while(r-l>=eps){
            double mid=(l+r)/2;
            if(check(mid)) final=mid,r=mid;
            else l=mid;
        }
        printf("%.2lf",final);
        //fclose(stdin);fclose(stdout);
        return 0;
    }
    /*
    6
    0 0
    10 0
    11 1
    15 1
    16 0
    25 0
    */
    
    

    模板题,不多说!先做映射,然后求下一个序列的最长不下降子序列!

    /*
        Name: P1439 【模板】最长公共子序列
        Copyright: njc 
        Author: Mudrobot
        Date: 2018/10/20 16:29:47
        Description: greedy!+Binary Search 
    */
    #include<bits/stdc++.h>
    #define gc() getchar()//caution!!!
    #define N 200000
    using namespace std;
    /*inline char gc() {
      static char buf[1<<18],*fs,*ft;
      return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<18,stdin)),fs==ft)?EOF:*fs++;
    }*/
    template<class T>
    inline void read(T &aa) {
      register int k=0,f=1;
      register char c=gc();
      for (;!isdigit(c);c=gc()) if(c=='-')f=-1;
      for (;isdigit(c);c=gc()) k=(k<<3)+(k<<1)+(c-'0');
      aa=k*f;
    }
    template<class T>
    inline void out(T x){if(x>9)out(x/10);putchar(x%10+'0');}
    struct sd{int x,y;}stk[N];
    int n,mp[N],nas[N],ma[N];
    bool cmp(sd a,sd b){if(a.x>b.x) return true;if(a.x==b.x&&a.y>b.y)return true;return false;}
    
    int main()
    {
        read(n);
        for(int i=1;i<=n;++i) read(stk[i].x),ma[stk[i].x]=i;
        for(int i=1;i<=n;++i) read(stk[i].y);//scanf("%d%d",&stk[i].x,&stk[i].y);
        for(int i=1;i<=n;++i) stk[i].y=ma[stk[i].y];
        int ans=0;
        for(int i=1;i<=n;++i)
        {
            if(stk[i].y>mp[ans])mp[++ans]=stk[i].y,nas[ans]++;
            else
            {
                int l=1,r=ans;
                while(l<=r)
                {
                    int mid=(l+r)/2;
                    if(mp[mid]>=stk[i].y)r=mid-1;
                    else l=mid+1;
                }
                mp[l]=stk[i].y;nas[l]++;
            }
        }
        int ns=0;
        for(int i=1;i<=ans;++i) ns=max(ns,nas[i]);
        printf("%d",max(ns,ans));
        return 0;
    }
    /*
    5 
    3 2 1 4 5
    1 2 3 4 5
    */
    
    

    注意 long double !!!

    /*
        Name: P1542 包裹快递
        Copyright: njc
        Author: Mudrobot
        Date: 2018/10/20 16:50:06
        Description: Binary Search
    */
    #include<bits/stdc++.h>
    #define gc() getchar()//caution!!!
    #define N 200005
    #define eps 0.000000001
    using namespace std;
    /*inline char gc() {
      static char buf[1<<18],*fs,*ft;
      return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<18,stdin)),fs==ft)?EOF:*fs++;
    }*/
    template<class T>
    inline void read(T &aa) {
      register int k=0,f=1;
      register char c=gc();
      for (;!isdigit(c);c=gc()) if(c=='-')f=-1;
      for (;isdigit(c);c=gc()) k=(k<<3)+(k<<1)+(c-'0');
      aa=k*f;
    }
    template<class T>
    inline void out(T x){if(x>9)out(x/10);putchar(x%10+'0');}
    struct sd{
        double len,s,e;
    }p[N];
    int n;
    bool check(long double v){
        long double tim=0;
        for(int i=1;i<=n;++i){
            tim+=(long double)p[i].len/v;
            //if(tim>(long double)p[i].s&&tim<(long double)p[i].e) continue;
            if(tim>(long double)p[i].e) return false;
            if(tim<(long double)(p[i].s))tim=p[i].s;
        }
        return true;
    }
    int main()
    {
        //freopen(".in", "r", stdin);freopen(".out", "w", stdout);
        read(n);
        for(int i=1;i<=n;++i) scanf("%lf%lf%lf",&p[i].s,&p[i].e,&p[i].len);
        long double l=0.0000000,r=0x7fffffff;
        long double final;
        while(r-l>=eps){
        	long double mid=(l+r)/2.00;
        	if(check(mid)) final=mid,r=mid;
        	else l=mid;
        }
        printf("%.2lf",(double)final);
        //fclose(stdin);fclose(stdout);
        return 0;
    }
    /*
    3
    1 2 2
    6 6 2
    7 8 4
    */
    
    
    
    #include<cstdio>
    #include<cstring>
    using namespace std;
    double a,b,c,d;
    double njc(double x)
    {
        return (a*x*x*x+b*x*x+c*x+d);
    }
    int main()
    {
        scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
        double p1,p2,tt;
        for(int i=-100;i<=100;++i)
        {
            p1=i; p2=i+1;
            if(njc(p1)==0) printf("%.2lf ",p1);
            else if(njc(p1)*njc(p2)<0)
            {
                while(p2-p1>=0.001)
                {
                    tt=(p2+p1)/2;
                    if(njc(p1)*njc(tt)<=0)
                    {
                        p2=tt;
                    }
                    else p1=tt;
                }
                printf("%.2lf ",p2);
            }
        }
    }
    
    /*
        Name: P1316 丢瓶盖
        Copyright: njc
        Author: Mudrobot
        Date: 2018/10/16 20:33:24
        Description: Binary_Search
    */
    #include<bits/stdc++.h>
    #define gc() getchar()//caution!!!
    #define N 100005
    using namespace std;
    /*inline char gc() {
      static char buf[1<<18],*fs,*ft;
      return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<18,stdin)),fs==ft)?EOF:*fs++;
    }*/
    template<class T>
    inline void read(T &aa) {
      register int k=0,f=1;
      register char c=gc();
      for (;!isdigit(c);c=gc()) if(c=='-')f=-1;
      for (;isdigit(c);c=gc()) k=(k<<3)+(k<<1)+(c-'0');
      aa=k*f;
    }
    template<class T>
    inline void out(T x){if(x>9)out(x/10);putchar(x%10+'0');}
    int n,k,a[N],minn=N*100,maxx=0;
    bool check(int kk){
        int cnt=1,s=1;
        for(int i=1;i<=n;++i){
            if(a[i]-a[s]<kk) continue;
            s=i;cnt++;
        }
        if(cnt>=k) return true;
        return false;
    }
    int main()
    {
        //freopen(".in", "r", stdin);freopen(".out", "w", stdout);
        read(n);read(k);
        for(int i=1;i<=n;++i) read(a[i]),maxx=max(maxx,a[i]),minn=min(minn,a[i]);
        sort(a+1,a+1+n);
        int l=1,r=maxx,ans;
        while(l<=r){
            int mid=(l+r)>>1;
            if(check(mid)) ans=mid,l=mid+1;
            else r=mid-1;
        }
        printf("%d",ans);
        //fclose(stdin);fclose(stdout);
        return 0;
    }
    /*
    5 3
    1 2 3 4 5
    */
    
    
    /*
        Name: P2370 yyy2015c01的U盘
        Copyright: njc
        Author: Mudrobot
        Date: 2018/10/17 21:05:24
        Description: Binary_Search
    */
    #include<bits/stdc++.h>
    #define gc() getchar()//caution!!!
    #define N 1005
    using namespace std;
    /*inline char gc() {
      static char buf[1<<18],*fs,*ft;
      return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<18,stdin)),fs==ft)?EOF:*fs++;
    }*/
    template<class T>
    inline void read(T &aa) {
      register int k=0,f=1;
      register char c=gc();
      for (;!isdigit(c);c=gc()) if(c=='-')f=-1;
      for (;isdigit(c);c=gc()) k=(k<<3)+(k<<1)+(c-'0');
      aa=k*f;
    }
    template<class T>
    inline void out(T x){if(x>9)out(x/10);putchar(x%10+'0');}
    struct sd{
        int w,val;
    }doc[N];
    int n,p,s,bag[N],maxx;
    bool check(int lim){//最大文件的大小 
        memset(bag,0,sizeof(bag));
        for(int i=1;i<=n;++i){
            if(doc[i].w>lim) continue;
            for(int j=s;j>=0;--j){
                if((j==0||bag[j])&&j+doc[i].w<=s){
                    bag[j+doc[i].w]=max(bag[j+doc[i].w],bag[j]+doc[i].val);
                }
            }
        }
        int ans=0;
        for(int i=1;i<=s;++i) ans=max(ans,bag[i]);
        return ans>=p;
    }
    int main()
    {
        //freopen(".in", "r", stdin);freopen(".out", "w", stdout);
        read(n);read(p);read(s);//p是我们希望的价值 
        for(int i=1;i<=n;++i){
            read(doc[i].w);read(doc[i].val);
            maxx=max(maxx,doc[i].w);
        }
        int l=1,r=maxx;
        int ans;
        while(l<r){
            int mid=(l+r)>>1;
            if(check(mid)) r=mid;
            else l=mid+1;
        }
        if(check(r))printf("%d",l);
        else printf("No Solution!");
        //fclose(stdin);fclose(stdout);
        return 0;
    }
    /*
    4 5 6
    5 1
    5 2
    5 3
    1 1
    */
    
    
    #include<bits/stdc++.h>
    using namespace std;
    int l,m,n;
    int len[50005],dis[50005];
    int minx=1000000;
    int search(int mid)
    {
        int cnt=0;
        for(int i=0;i<=n;++i)
        {
            int p=1;
            while(len[i+p]-len[i]<mid&&i+p<=n)
            {p++;cnt++;}
            i=i+p-1;
            
        }
        return cnt;
    }
    int main()
    {
        len[0]=0;
        scanf("%d%d%d",&l,&n,&m );
        if(m==0) {printf("%d",l);exit(0);} 
        for(int i=1;i<=n;++i)
        {
            scanf("%d",&len[i]);
            if(minx>len[i]) minx=len[i];
            dis[i-1]=len[i]-len[i-1];
        }
        len[n+1]=l;
        dis[n]=len[n+1]-len[n];
        int p1,p2;
        p1=0;p2=len[n];
        int mid;
        int ans;
        while(p1<=p2)
        {
            mid=(p1+p2)/2;
            int goal=search(mid);
            if(goal<=m){p1=mid+1;ans=mid;}
            if(goal>m){p2=mid-1;}
        }
        printf("%d",ans);
        return 0;
    } 
    
    /*
        Name: P1404 平均数
        Copyright: njc
        Author: Mudrobot
        Date: 2018/10/22 14:29:08
        Description: Binary Search
    */
    #include<bits/stdc++.h>
    #define N 100005
    typedef long long ll;
    using namespace std;
    ll n,m,s[N];
    double ans=0.0;
    ll q[N],t,h;   // 队列
    double k(ll x,ll y){  // 计算s[x],s[y]的斜率
        return (s[y]-s[x]+0.0)/(y-x);
    }
    int main() {
        cin>>n>>m;
        for (ll i=1,x;i<=n;i++){
            cin>>x; s[i]=s[i-1]+x;
        }
    
        for (ll i=m;i<=n;i++){
            while (t-h>=2 && k(i-m,q[t-1])<k(i-m,q[t-2])) t--;   // 删除上凸点
            q[t++]=i-m;  // 入队
            while (t-h>=2 && k(i,q[h])<k(i,q[h+1])) h++;  // 移动最大斜率点
            ans=max(ans,k(i,q[h]));
        }
        cout<<(ll)floor(ans*1000)<<endl;
        return 0;
    }
    /*
    10 6
    6
    4
    2
    10
    3
    8
    5
    9
    4
    1
    */
    

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  • 原文地址:https://www.cnblogs.com/mudrobot/p/13328974.html
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