[SHOI2008]堵塞的交通

Posted 2021-01-10 yicongli

神仙给我推荐的神仙题
线段树(6)个值维护联通性
然后查询时还有(4)种情况

不知道调试了多久

#include<bits/stdc++.h>

using namespace std;

#define gc c=getchar()
#define r(x) read(x)
#define ls (rt<<1)
#define rs (rt<<1|1)

template<typename T>
inline void read(T&x){
    x=0;T k=1;char gc;
    while(!isdigit(c)){if(c==‘-‘)k=-1;gc;}
    while(isdigit(c)){x=x*10+c-‘0‘;gc;}x*=k;
}

const int N=100000;

bool E[N][3];

struct seg{
    bool f[6];
    
    inline const bool& operator [](const int &x)const{
        return f[x];
    }
    
    inline bool& operator [](const int &x){
        return f[x];
    }
    
    inline void init(int x){
        static bool g[4][4];
        memset(g,0,sizeof(g));
        if(E[x][0])g[0][1]=g[1][0]=1;
        if(E[x][1])g[0][2]=g[2][0]=1;
        if(E[x][2])g[1][3]=g[3][1]=1;
        if(E[x+1][0])g[2][3]=g[3][2]=1;
        for(int k=0;k<4;++k){
            for(int i=0;i<4;++i){
                for(int j=0;j<4;++j){
                    g[i][j]|=g[i][k]&&g[k][j];
                }
            }
        }
        f[0]=g[0][1];
        f[1]=g[0][2];
        f[2]=g[1][3];
        f[3]=g[2][3];
        f[4]=g[0][3];
        f[5]=g[1][2];
    }
    
    inline friend seg operator + (const seg &a,const seg &b){
        seg ret;
        ret[0]=a[0]||(a[1]&&b[0]&&a[2]);
        ret[1]=(a[1]&&b[1])||(a[4]&&b[5]);
        ret[2]=(a[2]&&b[2])||(a[5]&&b[4]);
        ret[3]=b[3]||(b[1]&&a[3]&&b[2]);
        ret[4]=(a[4]&&b[2])||(a[1]&&b[4]);
        ret[5]=(a[5]&&b[1])||(a[2]&&b[5]);
        return ret;
    }
}tr[N<<2];

void modify(int rt,int l,int r,int x){
    if(l==r){
        tr[rt].init(l);
        return;
    }
    int mid=(l+r)>>1;
    if(x<=mid)modify(ls,l,mid,x);
    else modify(rs,mid+1,r,x);
    tr[rt]=tr[ls]+tr[rs];
}

seg query(int rt,int l,int r,int x,int y){
    if(x<=l&&r<=y)return tr[rt];
    int mid=(l+r)>>1;
    if(y<=mid)return query(ls,l,mid,x,y);
    else if(x>mid)return query(rs,mid+1,r,x,y);
    else return query(ls,l,mid,x,y)+query(rs,mid+1,r,x,y);
}

inline int get(bool x,bool y){
    if(x==0&&y==0)return 1;
    if(x==0&&y==1)return 4;
    if(x==1&&y==0)return 5;
    if(x==1&&y==1)return 2;
}

int n;
char s[10];

seg Tmp[2];

inline seg Query(int a,int b){
    return (a<b)?query(1,1,n,a,b-1):Tmp[E[a][0]];
}

int main(){
    r(n);
    Tmp[0][1]=Tmp[0][2]=1;
    memset(Tmp[1].f,1,sizeof(Tmp[1].f));
    while(1){
        scanf("%s",s);
        if(s[0]==‘E‘)break;
        int a,b,c,d;r(a),r(b),r(c),r(d);--a,--c;
        if(b>d)swap(a,c),swap(b,d);
        if(s[0]==‘A‘){
            seg l=Query(1,b),m=Query(b,d),r=Query(d,n);
            puts((m[get(a,c)]||((l[3]||m[0])&&m[get(a^1,c)])||((r[0]||m[3])&&m[get(a,c^1)])||((l[3]||m[0])&&(r[0]||m[3])&&m[get(a^1,c^1)]))?"Y":"N");
        }
        else {
            if(b==d)E[b][0]=(s[0]==‘O‘);
            else E[b][a+1]=(s[0]==‘O‘);
            if(b==d&&b>1)modify(1,1,n,b-1);
            modify(1,1,n,b);
        }
    }
}