bzoj2738 矩阵乘法 (整体二分)

题目链接：点这里！！！！

题意：就是求子矩阵的第K大。

题解：整体二分答案，然后利用二维树状数组维护下就可以了。和hdu5412很像。

hdu5412

1、无修改，你可以把值当成插入。

代码：

#include<cstdio>#include<cstring>#include<iostream>#include<sstream>#include<algorithm>#include<vector>#include<bitset>#include<set>#include<queue>#include<stack>#include<map>#include<cstdlib>#include<cmath>#define PI 2*asin(1.0)#define LL long long#define pb push_back#define pa pair<int,int>#define clr(a,b) memset(a,b,sizeof(a))#define lson lr<<1,l,mid#define rson lr<<1|1,mid+1,r#define bug(x) printf("%d++++++++++++++++++++%d\n",x,x)#define key_value ch[ch[root][1]][0]C:\Program Files\Git\binconst int  MOD = 1000000007;const int N = 500+15;const int maxn = 330000+1000;const int letter = 130;const int INF = 1e17;const double pi=acos(-1.0);const double eps=1e-8;using namespace std;inline int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}int n,q;int p[N][N],tot,max1;int ans[maxn],p1[maxn],p2[maxn];struct node{    int id,a,b,c,d;    int val;}qu[maxn];int lowbit(int x){return x&(-x);}void update(int i,int j,int val){    for(int x=i;x<=n;x+=lowbit(x))    for(int y=j;y<=n;y+=lowbit(y))        p[x][y]+=val;}int query(int i,int j){    int ans=0;    for(int x=i;x>0;x-=lowbit(x))    for(int y=j;y>0;y-=lowbit(y))        ans+=p[x][y];    return ans;}void solve(int L,int R,int low,int high){    if(L>R) return;    int i,j,sum;    if(low==high){        while(L<=R){            int j=p1[L];            if(qu[j].id==2) ans[j]=low;            L++;        }        return;    }    int mid=(low+high)/2;    int l=L,r=R;    for(i=L;i<=R;i++){        j=p1[i];        if(qu[j].id==2){            sum=query(qu[j].c,qu[j].d)+query(qu[j].a-1,qu[j].b-1)-query(qu[j].c,qu[j].b-1)-query(qu[j].a-1,qu[j].d);            if(sum<qu[j].val) qu[j].val-=sum,p2[r--]=j;            else p2[l++]=j;        }        else {            if(qu[j].val<=mid) update(qu[j].a,qu[j].b,1),p2[l++]=j;            else p2[r--]=j;        }    }    for(i=L;i<=R;i++){        j=p1[i];        if(qu[j].id==1&&qu[j].val<=mid) update(qu[j].a,qu[j].b,-1);    }    for(i=L;i<l;i++)p1[i]=p2[i];    for(r=R;i<=R;r--,i++)p1[i]=p2[r];    solve(L,l-1,low,mid);    solve(l,R,mid+1,high);}int main(){    n=read(),q=read();    max1=0;    for(int i=1;i<=n;i++)    for(int j=1;j<=n;j++){        qu[++tot].id=1;        qu[tot].a=i,qu[tot].b=j,qu[tot].val=read();        max1=max(max1,qu[tot].val);    }    while(q--){        qu[++tot].id=2;        qu[tot].a=read(),qu[tot].b=read();        qu[tot].c=read(),qu[tot].d=read();        qu[tot].val=read();    }    for(int i=1;i<=tot;i++) p1[i]=i;    solve(1,tot,0,max1);    for(int i=1;i<=tot;i++){        if(qu[i].id==2) printf("%d\n",ans[i]);    }    return 0;}

2、排好序在插入，就无需清空树状数组了，时间而且更快。

#include<cstdio>#include<cstring>#include<iostream>#include<sstream>#include<algorithm>#include<vector>#include<bitset>#include<set>#include<queue>#include<stack>#include<map>#include<cstdlib>#include<cmath>#define PI 2*asin(1.0)#define LL long long#define pb push_back#define pa pair<int,int>#define clr(a,b) memset(a,b,sizeof(a))#define lson lr<<1,l,mid#define rson lr<<1|1,mid+1,r#define bug(x) printf("%d++++++++++++++++++++%d\n",x,x)#define key_value ch[ch[root][1]][0]C:\Program Files\Git\binconst int  MOD = 1000000007;const int N = 500+15;const int maxn = 60000+1000;const int letter = 130;const int INF = 1e17;const double pi=acos(-1.0);const double eps=1e-8;using namespace std;inline int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}int n,q,cnt=0,max1,vs;int p[N][N];int p1[maxn],p2[maxn],ans[maxn];struct node{    int a,b,c,d,val;}qu[maxn],ps[N*N];int lowbit(int x){return x&(-x);}bool cmp(node a,node b){return a.val<b.val;}void update(int i,int j,int val){    for(int x=i;x<=n;x+=lowbit(x))    for(int y=j;y<=n;y+=lowbit(y))        p[x][y]+=val;}int query(int i,int j){    int ans=0;    for(int x=i;x>0;x-=lowbit(x))    for(int y=j;y>0;y-=lowbit(y))        ans+=p[x][y];    return ans;}void solve(int L,int R,int low,int high){    if(L>R) return;    int i,j,mid=(low+high)>>1,sum;    if(low==high){        while(L<=R){ans[p1[L]]=low,L++;}        return;    }    while(ps[vs+1].val<=mid) update(ps[vs+1].a,ps[vs+1].b,1),vs++;    while(ps[vs].val>mid) update(ps[vs].a,ps[vs].b,-1),vs--;    int l=L,r=R;    for(i=L;i<=R;i++){        j=p1[i];        sum=query(qu[j].c,qu[j].d)+query(qu[j].a-1,qu[j].b-1)-query(qu[j].c,qu[j].b-1)-query(qu[j].a-1,qu[j].d);        if(sum<qu[j].val) p2[r--]=j;        else p2[l++]=j;    }    for(i=L;i<l;i++) p1[i]=p2[i];    for(r=R;i<=R;r--,i++) p1[i]=p2[r];    solve(L,l-1,low,mid);    solve(l,R,mid+1,high);}int main(){    max1=0;    n=read(),q=read();    for(int i=1;i<=n;i++)    for(int j=1;j<=n;j++){        ps[++cnt].val=read();        ps[cnt].a=i,ps[cnt].b=j;        max1=max(max1,ps[cnt].val);    }    vs=0;    sort(ps+1,ps+cnt+1,cmp);    for(int i=1;i<=q;i++){        qu[i].a=read(),qu[i].b=read();        qu[i].c=read(),qu[i].d=read();        qu[i].val=read();    }    for(int i=1;i<=q;i++) p1[i]=i;    solve(1,q,0,max1);    for(int i=1;i<=q;i++) printf("%d\n",ans[i]);    return 0;}