hdu3911 Black And White(線段樹區間合并)

標籤：hdu   algorithm   線段樹   

題意：給一個由0,1組成的序列，有兩種操作，一種是翻轉給定區間的數(0->1,1->0)，另一種是查詢給定區間內由1組成的子串的最大長度。重點在區間合并和延遲標記。

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<queue>#include<set>#include<map>#include<vector>#include<cmath>#define INF 0x3fffffffusing namespace std;const int N=100010;struct Node{    int l,r,lsum0,rsum0,msum0,lsum1,rsum1,msum1;    int flag;//0不用翻轉，1用    int Mid(){        return (l+r)/2;    }}tree[4*N];int a[N];//合并左右子樹void pushUp(int rt){    int ll=tree[rt*2].r-tree[rt*2].l+1;    int rl=tree[rt*2+1].r-tree[rt*2+1].l+1;    //1的從左開始最大連續數    tree[rt].lsum1=tree[rt*2].lsum1;    //若等於左子樹長度，就要加上右子樹左邊最大    if(tree[rt*2].lsum1==ll){        tree[rt].lsum1+=tree[rt*2+1].lsum1;    }    //同理，1的從右開始最大連續數    tree[rt].rsum1=tree[rt*2+1].rsum1;    if(tree[rt*2+1].rsum1==rl){        tree[rt].rsum1+=tree[rt*2].rsum1;    }    //此區間的最大子串要麼在中間，要麼從左開始，要麼從右開始，所以最大子串為三種情況最大值。    tree[rt].msum1=max(max(tree[rt*2].msum1,tree[rt*2+1].msum1),tree[rt*2].rsum1+tree[rt*2+1].lsum1);    tree[rt].lsum0=tree[rt*2].lsum0;    if(tree[rt*2].lsum0==ll){        tree[rt].lsum0+=tree[rt*2+1].lsum0;    }    tree[rt].rsum0=tree[rt*2+1].rsum0;    if(tree[rt*2+1].rsum0==rl){        tree[rt].rsum0+=tree[rt*2].rsum0;    }    tree[rt].msum0=max(max(tree[rt*2].msum0,tree[rt*2+1].msum0),tree[rt*2].rsum0+tree[rt*2+1].lsum0);}//交換void Swap(int rt){    swap(tree[rt].lsum0,tree[rt].lsum1);    swap(tree[rt].rsum0,tree[rt].rsum1);    swap(tree[rt].msum0,tree[rt].msum1);}//從父節點向下更新(lazy)void pushDown(int rt){    if(tree[rt].flag==1){        tree[rt*2].flag^=1;        tree[rt*2+1].flag^=1;        tree[rt].flag=0;        Swap(rt*2);        Swap(rt*2+1);    }}void build(int rt,int l,int r){    tree[rt].l=l;tree[rt].r=r;tree[rt].flag=0;    //初始化每個葉子節點    if(l==r){        if(a[l]==0){            tree[rt].lsum0=tree[rt].rsum0=tree[rt].msum0=1;            tree[rt].lsum1=tree[rt].rsum1=tree[rt].msum1=0;        }        else if(a[l]==1){            tree[rt].lsum0=tree[rt].rsum0=tree[rt].msum0=0;            tree[rt].lsum1=tree[rt].rsum1=tree[rt].msum1=1;        }        return;    }    int mid=(l+r)/2;    build(2*rt,l,mid);    build(2*rt+1,mid+1,r);    pushUp(rt);}void update(int rt,int l,int r){    if(tree[rt].l==l&&tree[rt].r==r){        tree[rt].flag^=1;        Swap(rt);        return;    }    //更新子樹    pushDown(rt);    if(r<=tree[rt].Mid()){        update(2*rt,l,r);    }    else if(l>tree[rt].Mid()){        update(2*rt+1,l,r);    }    else{        update(2*rt,l,tree[rt].Mid());        update(2*rt+1,tree[rt].Mid()+1,r);    }    //向上合并更新    pushUp(rt);}int query(int rt,int l,int r){   if(tree[rt].l==l&&tree[rt].r==r) {        return tree[rt].msum1;   }   pushDown(rt);   int ans;   if(r<=tree[rt].Mid()){        ans=query(rt*2,l,r);   }   else if(l>tree[rt].Mid()){        ans=query(2*rt+1,l,r);   }   else{        int lr=query(2*rt,l,tree[rt].Mid());        int rr=query(2*rt+1,tree[rt].Mid()+1,r);        int a=tree[rt*2].rsum1;        if(a>tree[rt*2].r-l+1) a=tree[rt*2].r-l+1;        int b=tree[rt*2+1].lsum1;        if(b>r-tree[rt*2+1].l+1) b=r-tree[rt*2+1].l+1;        ans=max(max(lr,rr),a+b);   }   pushUp(rt);   return ans;}int main(){    //freopen("d:\\Test.txt","r",stdin);    int n;    while(scanf("%d",&n)!=EOF){        for(int i=1;i<=n;i++) scanf("%d",&a[i]);        build(1,1,n);        int m;        scanf("%d",&m);        while(m--){            int op,l,r;            scanf("%d%d%d",&op,&l,&r);            if(op==0){                cout<<query(1,l,r)<<endl;            }            else if(op==1){                update(1,l,r);            }        }    }    return 0;}

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hdu3911 Black And White(線段樹區間合并)