1926: [Sdoi2010]粟粟的书架 主席树+二分答案

一边看AlphaGo VS 李世石一边写这个sb题居然1A辣。
做法显然。n=1时就是序列上主席树然后二分答案，否则就是区间维护1000个前缀和暴力再二分答案。最后要暴力一下最小值的出现次数。

#include<iostream>#include<cstdio>#define ll long long #define M 10000005#define N 205using namespace std;int n,m,cnt,Q;int sum[M];int size[M],ls[M],rs[M];int s[1005][N][N];int ss[1005][N][N];int a[500005],b[N][N],v[1005];int root[500005];inline int read(){    int a=0,f=1; char c=getchar();    while (c<'0'||c>'9') {if (c=='-') f=-1; c=getchar();}    while (c>='0'&&c<='9') {a=a*10+c-'0'; c=getchar();}    return a*f;}void update(int l,int r,int x,int &y,int val){    y=++cnt;    ls[y]=ls[x]; rs[y]=rs[x]; sum[y]=sum[x]+val; size[y]=size[x]+1;    if (l==r) return;    int mid=l+r>>1;    if (val<=mid) update(l,mid,ls[x],ls[y],val);    else update(mid+1,r,rs[x],rs[y],val);}int find(int l,int r,int x,int y,int val){    int mid=l+r>>1;    if (l==r) return sum[y]-sum[x];    if (val<=mid) return find(l,mid,ls[x],ls[y],val)+sum[rs[y]]-sum[rs[x]];    else return find(mid+1,r,rs[x],rs[y],val);}int get_size(int l,int r,int x,int y,int val){    int mid=l+r>>1;    if (l==r) return size[y]-size[x];    if (val<=mid) return get_size(l,mid,ls[x],ls[y],val)+size[rs[y]]-size[rs[x]];    else return get_size(mid+1,r,rs[x],rs[y],val);}inline void solve1(){    for (int i=1;i<=m;i++) a[i]=read(),v[a[i]]++;    for (int i=1;i<=m;i++)        update(1,1000,root[i-1],root[i],a[i]);    while (Q--)    {        int x,l,r,H;        x=read(); l=read(); x=read(); r=read(); H=read();        int L=1,R=1000,ans=-1,last=0;        while (L<=R)        {            int mid=L+R>>1;            int tmp=find(1,1000,root[l-1],root[r],mid);            if (tmp>=H) last=tmp,ans=mid,L=mid+1; else R=mid-1;        }        if (ans==-1) {puts("Poor QLW"); continue;}        int now=get_size(1,1000,root[l-1],root[r],ans),rest=v[ans];        while (rest&&last-ans>=H) rest--,now--,last-=ans;        printf("%d\n",now);    }}int calc(int mid,int x1,int y1,int x2,int y2){    return ss[mid][x2][y2]+ss[mid][x1-1][y1-1]-ss[mid][x1-1][y2]-ss[mid][x2][y1-1];}int get_size1(int mid,int x1,int y1,int x2,int y2){    return s[mid][x2][y2]+s[mid][x1-1][y1-1]-s[mid][x1-1][y2]-s[mid][x2][y1-1];}inline void solve2(){    for (int i=1;i<=n;i++)        for (int j=1;j<=m;j++)            b[i][j]=read(),v[b[i][j]]++;    for (int k=1;k<=1000;k++)        for (int i=1;i<=n;i++)            for (int j=1;j<=m;j++)                s[k][i][j]=s[k][i-1][j]+s[k][i][j-1]-s[k][i-1][j-1]+(b[i][j]==k?1:0),ss[k][i][j]=s[k][i][j]*k;    for (int k=1000;k;k--)        for (int i=1;i<=n;i++)            for (int j=1;j<=m;j++)                ss[k][i][j]+=ss[k+1][i][j],s[k][i][j]+=s[k+1][i][j];    while (Q--)    {        int x1=read(),y1=read(),x2=read(),y2=read(),H=read();        int L=1,R=1000,ans=-1,last=0;        while (L<=R)        {            int mid=L+R>>1;            int tmp=calc(mid,x1,y1,x2,y2);            if (tmp>=H) last=tmp,ans=mid,L=mid+1; else R=mid-1;        }        if (ans==-1) {puts("Poor QLW"); continue;}        int now=get_size1(ans,x1,y1,x2,y2),rest=v[ans];        while (rest&&last-ans>=H) rest--,now--,last-=ans;        printf("%d\n",now);    }}int main(){    n=read(); m=read(); Q=read();    if (n==1) solve1(); else solve2();    return 0;}