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[BZOJ4199][NOI2015]品酒大會

-- d+ set space noi ble 更新 mat new

bzoj
luogu

sol

按照\(Height[i]\)從大到小的順序合並\(SA[i]\)\(SA[i-1]\)
考慮合並兩個集合對答案的貢獻。
方案數會增加\(sz[x]*sz[y]\)
最大值可以被\(max(mx[x]*mx[y],mn[x]*mn[y])\)更新

code

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define ll long long
int gi()
{
    int x=0,w=1;char ch=getchar();
    while
((ch<'0'||ch>'9')&&ch!='-') ch=getchar(); if (ch=='-') w=0,ch=getchar(); while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar(); return w?x:-x; } const int N = 300005; char S[N]; int n,a[N],t[N],x[N],y[N],SA[N],Rank[N],Height[N]; int
id[N],fa[N],sz[N],mx[N],mn[N]; ll ans[2][N]; bool cmp(int i,int j,int k){return y[i]==y[j]&&y[i+k]==y[j+k];} void getSA() { int m=30; for (int i=1;i<=n;++i) ++t[x[i]=a[i]]; for (int i=1;i<=m;++i) t[i]+=t[i-1]; for (int i=n;i>=1;--i) SA[t[x[i]]--]=i; for (int k=1;k<=n;k<<=1
) { int p=0; for (int i=0;i<=m;++i) y[i]=0; for (int i=n-k+1;i<=n;++i) y[++p]=i; for (int i=1;i<=n;++i) if (SA[i]>k) y[++p]=SA[i]-k; for (int i=0;i<=m;++i) t[i]=0; for (int i=1;i<=n;++i) ++t[x[y[i]]]; for (int i=1;i<=m;++i) t[i]+=t[i-1]; for (int i=n;i>=1;--i) SA[t[x[y[i]]]--]=y[i]; swap(x,y); x[SA[1]]=p=1; for (int i=2;i<=n;++i) x[SA[i]]=cmp(SA[i],SA[i-1],k)?p:++p; if (p>=n) break; m=p; } for (int i=1;i<=n;++i) Rank[SA[i]]=i; for (int i=1,j=0;i<=n;++i) { if (j) --j; while (a[i+j]==a[SA[Rank[i]-1]+j]) ++j; Height[Rank[i]]=j; } } bool comp(int i,int j){return Height[i]>Height[j];} int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);} void Merge(int x,int y,int z) { x=find(x);y=find(y); ans[0][z]+=(ll)sz[x]*sz[y]; sz[x]+=sz[y]; ans[1][z]=max(ans[1][z],max((ll)mx[x]*mx[y],(ll)mn[x]*mn[y])); mx[x]=max(mx[x],mx[y]); mn[x]=min(mn[x],mn[y]); fa[y]=x; } int main() { n=gi();scanf("%s",S+1); for (int i=1;i<=n;++i) a[i]=S[i]-'a'+1; getSA(); for (int i=1;i<=n;++i) id[i]=fa[i]=i,sz[i]=1,mx[i]=mn[i]=gi(); sort(id+2,id+n+1,comp); memset(ans[1],-127,sizeof(ans[1])); for (int i=2;i<=n;++i) Merge(SA[id[i]],SA[id[i]-1],Height[id[i]]); for (int i=n-2;i>=0;--i) ans[0][i]+=ans[0][i+1],ans[1][i]=max(ans[1][i],ans[1][i+1]); for (int i=0;i<n;++i) printf("%lld %lld\n",ans[0][i],ans[0][i]?ans[1][i]:0); return 0; }

[BZOJ4199][NOI2015]品酒大會