【补题日记】[2022杭电暑期多校3]K-Taxi

Pro

https://acm.hdu.edu.cn/showproblem.php?pid=7172

Sol

先考虑没有w的限制的我不会的经典模型怎么做。

$$\because |x|=max(x,-x)$$
$$\therefore max(|x^{{}^{\prime }}-x_i|+|y^{{}^{\prime }}-y_i|)$$
= m a x { m a x ( x ′ − x i , − x ′ + x i ) + m a x ( y ′ − y i , − y ′ + y i ) } =max\{max(x^{'}-x_i, -x^{'}+x_i)+ {max(y^{'}-y_i, -y^{'}+y_i)}\} =max{max(x′−xi​,−x′+xi​)+max(y′−yi​,−y′+yi​)}
= m a x { x ′ − x i + y ′ − y i , x ′ − x i − y ′ + y i , − x ′ + x i + y ′ − y i , − x ′ + x i − y ′ + y i } =max\{x^{'}-x_i+y^{'}-y_i, x^{'}-x_i-y^{'}+y_i, -x^{'}+x_i+y^{'}-y_i,-x^{'}+x_i-y^{'}+y_i \} =max{x′−xi​+y′−yi​,x′−xi​−y′+yi​,−x′+xi​+y′−yi​,−x′+xi​−y′+yi​}
$$=max\{(x^{{}^{\prime }}+y^{{}^{\prime }})+(-x_i-y_i),(x^{{}^{\prime }}-y^{{}^{\prime }})+(-x_i+y_i),(-x^{{}^{\prime }}+y^{{}^{\prime }})+(x_i-y_i),(-x^{{}^{\prime }}-y^{{}^{\prime }})+(x_i+y_i)\}$$
因此只需要维护$-x_i-y_i,-x_i+y_i,x_i-y_i,x_i+y_i$就可以$O(1)$的时间求出距离该点的曼哈顿距离最大值。

而本题加入w的限制后，有以下两种二分思路来做：（经过测试，第一种虽做法虽然只比第二种多了个log，但是因为此处的n比较大，因此不能通过；第二种方法关掉流同步的cin和快读也不能A掉，只能scanf才A掉哈哈哈，归为hdu机子的问题上次1e5的O(n)都给我T了我还记着呢）

主要是check(judge)函数的区别：

1、直接二分答案，比如想要判断答案是不是大于等于x，那么找到第一个权值大于等于x的点，根据上面的式子求出原始模型的距离，然后与x比较即可。

因为外面有一层二分，里面可以使用lower_bound二分得到点的下标，因此总的时间复杂度为$O(nlognlogn)$

2、二分城镇
考虑判断第x个城镇，它的点权为a[x].w，求出编号大于等于x的城镇中到询问点的最大曼哈顿距离d

如果d>a[x].w，答案一定大于等于a[x].w，因此需要进一步二分右半区间；反之同理。总的时间复杂度为$O(nlogn)$

Code

1.二分答案：

//By cls1277
#include
using namespace std;
typedef long long LL;
#define Fo(i,a,b) for(LL i=(a); i<=(b); i++)
#define Ro(i,b,a) for(LL i=(b); i>=(a); i--)
#define Eo(i,x,_) for(LL i=head[x]; i; i=_[i].next)
#define Ms(a,b) memset((a),(b),sizeof(a))
#define endl '\n'

const LL maxn = 1e5+5;
LL n, q;
struct Node {
    LL x, y, w;
}a[maxn];
LL b[maxn];

inline LL read() {
	LL x = 0, f = 1;char c = getchar();
	while (!isdigit(c)) { if (c == '-')f = -f;c = getchar(); }
	while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48ll), c = getchar();
	return x * f;
}

LL f[4][maxn]; //0:x+y 1:x-y 2:-x+y 3:-x-y

bool operator < (const Node&x, const Node&y) {
    return x.w<y.w;
}

bool judge(LL x, LL y, LL w) {
    LL idx = lower_bound(b+1, b+n+1, w)-b;
    LL d = max({x+y+f[3][idx], x-y+f[2][idx], -x+y+f[1][idx], -x-y+f[0][idx]});
    return (d>=w);
}

int main() {
    // ios::sync_with_stdio(false);
    // cin.tie(nullptr);
    #ifdef DEBUG
    freopen("data.txt","r",stdin);
    freopen("out.txt","w",stdout);
    #endif
    LL t; scanf("%lld",&t); //t=read(); //cin>>t;
    while(t--) {
        scanf("%lld%lld",&n,&q);
        // n=read(); q=read(); //cin>>n>>q;
        Fo(i,1,n) {
            // a[i].x = read();
            // a[i].y = read();
            // a[i].w = read();
            scanf("%lld%lld%lld",&a[i].x,&a[i].y,&a[i].w);
        }
        //cin>>a[i].x>>a[i].y>>a[i].w;
        sort(a+1, a+n+1);
        Fo(i,1,n) {
            b[i] = a[i].w;
            f[0][i] = f[1][i] = f[2][i] = f[3][i] = 0;
        }
        f[0][n] = a[n].x+a[n].y;
        f[1][n] = a[n].x-a[n].y;
        f[2][n] = -a[n].x+a[n].y;
        f[3][n] = -a[n].x-a[n].y;
        Ro(i,n-1,1) {
            f[0][i] = max(f[0][i+1], a[i].x+a[i].y);
            f[1][i] = max(f[1][i+1], a[i].x-a[i].y);
            f[2][i] = max(f[2][i+1], -a[i].x+a[i].y);
            f[3][i] = max(f[3][i+1], -a[i].x-a[i].y);            
        }
        while(q--) {
            LL _x, _y; scanf("%lld%lld", &_x, &_y); // _x=read(); _y=read(); //cin>>_x>>_y;
            LL l=0, r=a[n].w;
            while(l<r) {
                LL mid=(l+r+1)>>1;
                if(judge(_x, _y, mid)) l = mid;
                else r = mid-1;
            }
            // cout<
            printf("%lld\n",l);
        }
    }
    return 0;
}
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2.二分城镇：

//By cls1277
#include
using namespace std;
typedef long long LL;
#define Fo(i,a,b) for(LL i=(a); i<=(b); i++)
#define Ro(i,b,a) for(LL i=(b); i>=(a); i--)
#define Eo(i,x,_) for(LL i=head[x]; i; i=_[i].next)
#define Ms(a,b) memset((a),(b),sizeof(a))
#define endl '\n'

const LL maxn = 1e5+5;
LL n, q;
struct Node {
    LL x, y, w;
}a[maxn];
LL b[maxn];

inline LL read() {
	LL x = 0, f = 1;char c = getchar();
	while (!isdigit(c)) { if (c == '-')f = -f;c = getchar(); }
	while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48ll), c = getchar();
	return x * f;
}

LL f[4][maxn]; //0:x+y 1:x-y 2:-x+y 3:-x-y

bool operator < (const Node&x, const Node&y) {
    return x.w<y.w;
}

LL ans = 0;

bool judge(LL x, LL y, LL idx) {
    LL d = max({x+y+f[3][idx], x-y+f[2][idx], -x+y+f[1][idx], -x-y+f[0][idx]});
    if(d>=a[idx].w) {
        ans = max(ans, a[idx].w);
        return true;
    } else {
        ans = max(ans, d);
        return false;
    }
}

int main() {
    // ios::sync_with_stdio(false);
    // cin.tie(nullptr);
    #ifdef DEBUG
    freopen("data.txt","r",stdin);
    freopen("out.txt","w",stdout);
    #endif
    LL t; scanf("%lld",&t); //t=read(); //cin>>t;
    while(t--) {
        scanf("%lld%lld",&n,&q);
        // n=read(); q=read(); //cin>>n>>q;
        Fo(i,1,n) {
            // a[i].x = read();
            // a[i].y = read();
            // a[i].w = read();
            scanf("%lld%lld%lld",&a[i].x,&a[i].y,&a[i].w);
        }
        //cin>>a[i].x>>a[i].y>>a[i].w;
        sort(a+1, a+n+1);
        Fo(i,1,n) {
            b[i] = a[i].w;
            f[0][i] = f[1][i] = f[2][i] = f[3][i] = 0;
        }
        f[0][n] = a[n].x+a[n].y;
        f[1][n] = a[n].x-a[n].y;
        f[2][n] = -a[n].x+a[n].y;
        f[3][n] = -a[n].x-a[n].y;
        Ro(i,n-1,1) {
            f[0][i] = max(f[0][i+1], a[i].x+a[i].y);
            f[1][i] = max(f[1][i+1], a[i].x-a[i].y);
            f[2][i] = max(f[2][i+1], -a[i].x+a[i].y);
            f[3][i] = max(f[3][i+1], -a[i].x-a[i].y);            
        }
        while(q--) {
            LL _x, _y; scanf("%lld%lld", &_x, &_y); // _x=read(); _y=read(); //cin>>_x>>_y;
            LL l=0, r=n;
            ans = 0;
            while(l<r) {
                LL mid=(l+r+1)>>1;
                if(judge(_x, _y, mid)) l = mid;
                else r = mid-1;
            }
            // cout<
            printf("%lld\n",ans);
        }
    }
    return 0;
}
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