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zc
2020-01-18 22:40:00
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我们考虑两个断点(设为l,r)符合的条件:

每种颜色的珠子只能出现在其中一条链中

也就是若[l,r]中出现了某种颜色[l,r]也就包含了所有的这种颜色

我们可以记录颜色出现的前缀和

记录前i个位置每种颜色的出现次数,如果位置i是 颜色a[i]的最后一个位置, 就把颜色 a[i] 的结果清零。

这样若两个位置的前缀和相同,就是合法的

可以通过hash来判断,记录所有hash值,排序后统计

两个分割点 l,r 要尽可能满足 r−l 接近 \frac n2

可以用类似双指针的方式维护

这题卡hash,要双hash

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i(x);i<=y;++i)
    #define Fdr(i,x,y) for(int i(x);i>=y;--i)
    #define fl(i,x) for(int i(head[x]),to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) ((int)(log2(n)))
    #define inf 2122219134
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const int str=1<<20;static char in_buf[str],*in_s,*in_t;bool __=0;il char gc(){return (in_s==in_t)&&(in_t=(in_s=in_buf)+fread(in_buf,1,str,stdin)),in_s==in_t?EOF:*in_s++;}il void in(string &ch){ch.clear();if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)__=1;}il void in(char &ch){if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)__=1;else ch=c;}il void in(char *ch){*ch='\0';if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)__=1;*ch='\0';}template<typename T>il void in(T &x){if(__)return;char c=gc();bool f=0;while(c!=EOF&&(c<'0'||c>'9'))f^=(c=='-'),c=gc();if(c==EOF){__=1;return;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=x*10+c-48,c=gc();if(c==EOF)__=1;if(f)x=-x;}template<typename T,typename ... arr>il void in(T &x,arr & ... y){in(x),in(y...);}const char ln='\n';static char out_buf[str],*out_s=out_buf,*out_t=out_buf+str;il void flush(){fwrite(out_buf,1,out_s-out_buf,stdout);out_s=out_buf;}il void pt(char c){(out_s==out_t)?(fwrite(out_s=out_buf,1,str,stdout),*out_s++=c):(*out_s++=c);}il void out(const char* s){while(*s)pt(*s++);}il void out(char* s){while(*s)pt(*s++);}il void out(char c){pt(c);}il void out(string s){for(int i=0;s[i];i++)pt(s[i]);}template<typename T>il void out(T x){if(!x){pt('0');return;}if(x<0)pt('-'),x=-x;char a[50],t=0;while(x)a[t++]=x%10,x/= 10;while(t--)pt(a[t]+'0');}template<typename T,typename ... arr>il void out(T x,arr & ... y){out(x),out(y...);}}using namespace IO;
#define N 1000011
int n,k,a[N],b[N],c[N],p1=133331,p2=23333,m1=100000007,m2=19260817,ans=inf;
ll s1[N],s2[N],cnt=0;
struct node{
    int id,s1,s2;
    bool operator<(node x)const{
        if(s1!=x.s1)return s1<x.s1;
        if(s2!=x.s2)return s2<x.s2;
        return id<x.id;
    }
}t[N];
int main(){
    in(n,k);
    s1[0]=s2[0]=1;
    Fur(i,1,n)
        in(a[i]),
        s1[i]=s1[i-1]*p1%m1,
        s2[i]=s2[i-1]*p2%m2;
    Fdr(i,n,1)if(!b[a[i]])b[a[i]]=i;
    ll sum1=0,sum2=0;
    Fur(i,1,n){
        ++c[a[i]];
        sum1=(sum1+s1[a[i]])%m1;
        sum2=(sum2+s2[a[i]])%m2;
        if(b[a[i]]==i)
            sum1=(sum1-s1[a[i]]*c[a[i]]%m1+m1)%m1,
            sum2=(sum2-s2[a[i]]*c[a[i]]%m2+m2)%m2;
        t[i]=node{i,sum1,sum2};
    }
    sort(t+1,t+n+1);
    int mid=(n+1)/2;
    for(int i=1;i<=n;){
        int nxt=i;
        while(nxt<=n&&t[nxt].s1==t[i].s1&&t[nxt].s2==t[i].s2)++nxt;
        cnt+=1ll*(nxt-i)*(nxt-i-1)/2;
        for(int l=i,r=i;r<nxt;++r){
            while(l<r&&t[r].id-t[l].id>=mid)++l;
            int tp=ABS(n-2*(t[r].id-t[l].id));
            if(l>i){
                int tmp=ABS(n-2*(t[r].id-t[l-1].id));
                tp=MIN(tp,tmp);
            }
            ans=MIN(ans,tp);
        }
        i=nxt;
    }
    cout<<cnt<<" "<<ans<<endl;
}
LG 3587 [POI2015]POD
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