bzoj1858: [Scoi2010]序列操作

lazy-tag线段树。

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 800000 + 10;

struct Segtree {
    #define lc(x) ((x)<<1)
    #define rc(x) (((x)<<1)|1)
    
    int l[maxn],r[maxn];
    int l1[maxn],r1[maxn],con1[maxn],cnt1[maxn];
    int l0[maxn],r0[maxn],con0[maxn];
    int t1[maxn],t0[maxn],tr[maxn];
    
    void set1(int x) {
        l1[x]=r1[x]=con1[x]=cnt1[x]=r[x]-l[x]+1;
        l0[x]=r0[x]=con0[x]=0;
        t1[x]=1; t0[x]=tr[x]=0;    
    }
    
    void set0(int x) {
        l0[x]=r0[x]=con0[x]=r[x]-l[x]+1;
        l1[x]=r1[x]=con1[x]=cnt1[x]=0;
        t0[x]=1; t1[x]=tr[x]=0;
    }
    
    void inv(int x) {
        swap(l1[x],l0[x]);
        swap(r1[x],r0[x]);
        swap(con1[x],con0[x]);
        cnt1[x]=r[x]-l[x]+1-cnt1[x];    
        tr[x]^=1;
    }
    
    void push(int x) {
        if(t0[x]) {
            set0(lc(x));
            set0(rc(x));
            t0[x]=0;    
        }
        if(t1[x]) {
            set1(lc(x));
            set1(rc(x));
            t1[x]=0;
        }
        if(tr[x]) {
            inv(lc(x));
            inv(rc(x));
            tr[x]=0;
        }
    }
    
    void update(int x) {
        l1[x]=(l1[lc(x)]==r[lc(x)]-l[lc(x)]+1?l1[lc(x)]+l1[rc(x)]:l1[lc(x)]);
        r1[x]=(r1[rc(x)]==r[rc(x)]-l[rc(x)]+1?r1[rc(x)]+r1[lc(x)]:r1[rc(x)]);
        l0[x]=(l0[lc(x)]==r[lc(x)]-l[lc(x)]+1?l0[lc(x)]+l0[rc(x)]:l0[lc(x)]);
        r0[x]=(r0[rc(x)]==r[rc(x)]-l[rc(x)]+1?r0[rc(x)]+r0[lc(x)]:r0[rc(x)]);    
        con1[x]=max(max(con1[lc(x)],con1[rc(x)]),r1[lc(x)]+l1[rc(x)]);
        con0[x]=max(max(con0[lc(x)],con0[rc(x)]),r0[lc(x)]+l0[rc(x)]);
        cnt1[x]=cnt1[lc(x)]+cnt1[rc(x)];
    }
    
    void op0(int x,int L,int R) {
        push(x);
        if(R<l[x] || L>r[x]) return;
        if(L<=l[x] && r[x]<=R) {
            set0(x);
            return;
        }
        op0(lc(x),L,R);
        op0(rc(x),L,R);
        update(x);
    }
    
    void op1(int x,int L,int R) {
        push(x);
        if(R<l[x] || L>r[x]) return;
        if(L<=l[x] && r[x]<=R) {
            set1(x);
            return;
        }
        op1(lc(x),L,R);
        op1(rc(x),L,R);
        update(x);
    }
    
    void op2(int x,int L,int R) {
        push(x);
        if(R<l[x] || L>r[x]) return;
        if(L<=l[x] && r[x]<=R) {
            inv(x);
            return;
        }
        op2(lc(x),L,R);
        op2(rc(x),L,R);
        update(x);
    }
    
    int op3(int x,int L,int R) {
        push(x);
        if(R<l[x] || L>r[x]) return 0;
        if(L<=l[x] && r[x]<=R) return cnt1[x];
        return(op3(lc(x),L,R)+op3(rc(x),L,R));
    }
    
    int op4(int x,int L,int R) {
        push(x);
        if(R<l[x] || L>r[x]) return 0;
        if(L<=l[x] && r[x]<=R) return con1[x];
        return max(max(op4(lc(x),L,R),op4(rc(x),L,R)),min(r1[lc(x)],r[lc(x)]-L+1)+min(l1[rc(x)],R-l[rc(x)]+1));
    }
    
    void build(int x,int L,int R) {
        l[x]=L; r[x]=R;
        if(L==R) {
            int k;
            scanf("%d",&k);
            if(k) l1[x]=r1[x]=con1[x]=cnt1[x]=1;
            else l0[x]=r0[x]=con0[x]=1;    
            return;
        }
        int mid=(L+R)>>1;
        build(lc(x),L,mid);
        build(rc(x),mid+1,R);
        update(x);    
    }
}seg;
int n,m;

int main() {
    scanf("%d%d",&n,&m);
    seg.build(1,0,n-1);
    for(int i=1,op,a,b;i<=m;i++) {
        scanf("%d%d%d",&op,&a,&b);
        if(op==0) seg.op0(1,a,b);
        else if(op==1) seg.op1(1,a,b);
        else if(op==2) seg.op2(1,a,b);
        else if(op==3) printf("%d\n",seg.op3(1,a,b));
        else if(op==4) printf("%d\n",seg.op4(1,a,b)); 
    }
    return 0;
}
posted @ 2016-05-15 20:07  invoid  阅读(121)  评论(0编辑  收藏  举报