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  • P1020 导弹拦截

    严格上升的话,减1就好啦

    #include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<queue>
    #include<algorithm>
    #include<cmath>
    #include<cstring>
    #include<sstream>
    #pragma GCC optimize(2)
    #define inf 2147483647
    #define N 1000010
    #define For(i,a,b) for(int i=a;i<=b;++i)
    #define p(a) putchar(a)
    #define g() getchar()
    
    using namespace std;
    int n,ans,m;
    int f[N],t[N],b[N];
    int a[N];
    string s;
    char c[100];
    void in(int &x){
        int y=1;char c=getchar();x=0;
        while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
        while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=getchar();}
        x*=y;
    }
    void o(int x){
        if(x<0){putchar('-');x=-x;}
        if(x>9)o(x/10);
        putchar(x%10+'0');
    }
    
    int getmax(int k){
        int Max=0;
        for(;k>0;k-=(-k)&k)
            Max=max(Max,t[k]);
        return Max;
    }
    
    void modify(int k,int Max){
        for(;k<=500000;k+=(-k)&k)
            t[k]=max(t[k],Max);
    }
    
    int main(){
        getline(cin,s);
        istringstream ss(s);
        while(ss>>c){
            int len=(int)strlen(c);
            n++;
            For(i,0,len-1)
                a[n]=(a[n]<<1)+(a[n]<<3)+c[i]-'0';
        }
          For(i,1,n)
              b[i]=a[n-i+1];
          For(i,1,n){
              f[i]=getmax(b[i])+1;
              ans=max(ans,f[i]);
              modify(b[i],f[i]);
          }
          o(ans);p('
    ');
          ans=0;
          memset(t,0,sizeof(t));
          memset(f,0,sizeof(f));
          For(i,1,n){
              f[i]=getmax(a[i]-1)+1;
              ans=max(ans,f[i]);
              modify(a[i],f[i]);
          }
          o(ans);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/war1111/p/11268198.html
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