掃描線專題

P1904 天際線

題目連接：https://www.luogu.com.cn/problem/P1904

 

想法：

咱們根據題目的意思不難發現，咱們能夠維護一個x軸區間對應的 y 值的最大值。而後咱們去遍歷咱們的 x 軸對應的 y 值，若是前面一個和當前的 y 相同就說明仍是直線，沒有出現轉折點。

可是出現了一個問題就是：爲了不邊界出現問題，因此咱們採起和掃描線同樣的作法，讓 【l，r】 表明 【l，r+1】

 

#include <cstdio>
#include <iostream>
#include <algorithm>

#define ll long long
#define ull unsigned long long
#define ls nod<<1
#define rs (nod<<1)+1
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define INF 0x3f3f3f3f
#define max(a, b) (a>b?a:b)
#define min(a, b) (a<b?a:b)


const double eps = 1e-8;
const int maxn = 1e6 + 10;
const ll MOD = 1e9 + 7;
const int mlog=20;

int sgn(double a) { return a < -eps ? -1 : a < eps ? 0 : 1; }

using namespace std;

struct node {
    int x,h,y;
    bool operator < (const node &a) const {
        return h < a.h;
    }
}q[maxn];

int ler[maxn],x[maxn],y[maxn];

struct segment_tree {
    int sum;
    int lazy;
}tree[maxn << 2];

void pushdown(int l,int r,int nod) {
    if (tree[nod].lazy) {
        tree[ls].lazy = tree[rs].lazy = tree[nod].lazy;
        tree[ls].sum = tree[rs].sum = tree[nod].sum;
        tree[nod].lazy = 0;
    }
}

void modify(int l,int r,int x,int y,int z,int nod) {
    if (x <= l && y >= r) {
        tree[nod].sum = z;
        tree[nod].lazy = 1;
        return ;
    }
    pushdown(l,r,nod);
    int mid = (l + r) >> 1;
    if (x <= mid)
        modify(l,mid,x,y,z,ls);
    if (y > mid)
        modify(mid+1,r,x,y,z,rs);
}

int query(int l,int r,int k,int nod) {
    if (l == r)
        return tree[nod].sum;
    int mid = (l + r) >> 1;
    pushdown(l,r,nod);
    if (k <= mid)
        return query(l,mid,k,ls);
    else
        return query(mid+1,r,k,rs);
}

int ef(int l1,int r1,int mb)
{
    int l=l1,r=r1;
    while (l<=r)
    {
        int mid=(l+r)>>1;
        if (ler[mid]>=mb)r=mid-1;
        else l=mid+1;
    }
    return l;
}

int main() {
#ifdef ONLINE_JUDGE
#else

    freopen("/Users/ackerman/CLionProjects/codeforces/in.txt","r",stdin);
#endif
    int m = 1;
    int len = 0;
    while (~scanf("%d%d%d",&q[m].x,&q[m].h,&q[m].y)) {
        ler[++len] = q[m].x;
        ler[++len] = q[m].y;
        m++;
    }
    m--;
    sort(ler+1,ler+1+len);
    sort(q+1,q+1+m);
    int n = 1;
    for (int i = 1;i < len;i++) {
        if (ler[i] != ler[i+1])
            ler[++n] = ler[i+1];
    }
    for (int i = 1;i <= m;i++) {
        q[i].x = ef(1,n,q[i].x);
        q[i].y = ef(1,n,q[i].y);
        modify(1,n,q[i].x,q[i].y-1,q[i].h,1);
    }
    for (int i = 1;i <= n;i++) {
        x[i] = ler[i];
        y[i] = query(1,n,i,1);
    }
    for (int i = 1;i <= n;i++) {
        if (y[i] != y[i-1]) {
            printf("%d ", x[i]);
            printf("%d ", y[i]);
        }
    }
    return 0;
}

 

P1502 窗口的星星

題目連接：https://www.luogu.com.cn/problem/P1502

 

想法：

由於給咱們的是一個固定的矩形。咱們不妨就讓每個星星其實就是一個以該星星爲矩形左下角的一個固定大小的矩陣，因爲邊界上的星星不能算，因此咱們分別把長和寬都進行-1的操做。

而後這樣的話問題就轉化爲矩形相交的問題，我要求的就是相交的那段區間裏面的最大亮度就能夠了。因此問題再次轉化爲掃描線的思想。可是這裏的話，因爲咱們已經對邊界進行了操做因此不會出現前面一道題的邊界的問題，因此咱們直接正常的線段樹維護就行了。

 

#include <cstdio>
#include <iostream>
#include <algorithm>

#define ll long long
#define ull unsigned long long
#define ls nod<<1
#define rs (nod<<1)+1
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define INF 0x3f3f3f3f
#define max(a, b) (a>b?a:b)
#define min(a, b) (a<b?a:b)


const double eps = 1e-8;
const int maxn = 1e6 + 10;
const ll MOD = 1e9 + 7;
const int mlog=20;

int sgn(double a) { return a < -eps ? -1 : a < eps ? 0 : 1; }

using namespace std;

ll x1,y1,x2,y2,X[maxn << 1];

struct ScanLine {
    ll l,r,h,mark;

    bool operator < (const ScanLine &rhs) const {
        if (h == rhs.h)
            return mark > rhs.mark;
        return h < rhs.h;
    }
} line[maxn << 2];

struct segment_tree {
    int l,r;
    ll sum,lazy;
}tree[maxn << 2];

void build(int l,int r,int nod) {
    tree[nod].l = l,tree[nod].r = r;
    tree[nod].lazy = 0;
    tree[nod].sum = 0;
    if (l == r)
        return ;
    int mid = (l + r) >> 1;
    build(l,mid,ls);
    build(mid+1,r,rs);
}

void push_up(int nod) {
    tree[nod].sum = max(tree[ls].sum,tree[rs].sum);
}

void pushdown(int nod) {
    if (tree[nod].lazy) {
        tree[ls].sum += tree[nod].lazy;
        tree[rs].sum += tree[nod].lazy;
        tree[ls].lazy += tree[nod].lazy;
        tree[rs].lazy += tree[nod].lazy;
        tree[nod].lazy = 0;
    }
}

void modify(ll x,ll y,int f,int nod) {
    int l = tree[nod].l,r = tree[nod].r;
    if (x <= l && y >= r) {
        tree[nod].sum += f;
        tree[nod].lazy += f;
        return ;
    }
    pushdown(nod);
    int mid = (l + r) >> 1;
    if (x <= mid)
        modify(x,y,f,ls);
    if (y > mid)
        modify(x,y,f,rs);
    push_up(nod);
}

int main() {
    int t;
    scanf("%d",&t);
    while (t--) {
        ll n,w,h;
        scanf("%lld%lld%lld",&n,&w,&h);
        for (int i = 1;i <= n;i++) {
            ll x,y,l;
            scanf("%lld%lld%lld",&x,&y,&l);
            X[2 * i - 1] = x;
            X[2 * i] = x + w - 1;
            line[2 * i - 1] = {x,x+w-1,y,l};
            line[2 * i] = {x,x+w-1,y+h-1,-l};
        }
        n <<= 1;
        sort(X + 1,X + 1 + n);
        sort(line + 1,line + 1 + n);
        ll tot = unique(X + 1,X + 1 + n) - X - 1;
        build(1,tot,1);
        ll ans = 0;
        for (int i = 1;i <= n;i++) {
            ll l = upper_bound(X + 1,X + 1 + tot,line[i].l) - X - 1;
            ll r = upper_bound(X + 1,X + 1 + tot,line[i].r) - X - 1;
            modify(l,r,line[i].mark,1);
            ans = max(ans,tree[1].sum);
        }
        printf("%lld\n",ans);
    }
    return 0;
}

 

 

P5816 [CQOI2010]內部白點

題目連接：https://www.luogu.com.cn/problem/P5816

 

想法：

 

（圖來自ez_lcw）

 

只有水平線和垂直線纔會對白點的出現產生影響，因此咱們能夠維護這兩種的線段。

若是出現了水平線，咱們就查詢這段區間。 若是出現了垂線，下端點咱們讓他+1，上端點咱們讓他-1

而後按照 y 的順序對這些線段進行排序。

 

#include <cstdio>
#include <iostream>
#include <algorithm>

#define ll long long
#define ull unsigned long long
#define ls nod<<1
#define rs (nod<<1)+1
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define INF 0x3f3f3f3f
#define max(a, b) (a>b?a:b)
#define min(a, b) (a<b?a:b)


const double eps = 1e-8;
const int maxn = 1e5 + 10;
const ll MOD = 1e9 + 7;
const int mlog=20;

int sgn(double a) { return a < -eps ? -1 : a < eps ? 0 : 1; }

using namespace std;

struct BIT {
    ll c[maxn << 2]; int n;

    void init(int sz) {n=sz;for(int i = 0;i <= n+1;i++) c[i]=0;}

    ll qwq(int x) {
        ll ans=0; for(;x;x-=x&-x) ans+=c[x];
        return ans;
    }
    ll qry(int l,int r) {return qwq(r)-qwq(l-1);}
    void upd(int x,ll v=1) {if(x!=0) for(;x<=n;x+=x&-x) c[x]+=v;}
}bit;

struct ScanLine {
    ll l,r,h;
    int mark;

    bool operator < (const ScanLine &rhs) const {
        if (h == rhs.h)
            return mark < rhs.mark;
        return h < rhs.h;
    }
} line[maxn << 2];

struct node {
    int x,y;
}p[maxn];

int x[maxn];

bool cmpx(node a,node b) {
    if (a.x == b.x)
        return a.y < b.y;
    return a.x < b.x;
}

bool cmpy(node a,node b) {
    if (a.y == b.y)
        return a.x < b.x;
    return a.y < b.y;
}

int main() {
    int n;
    scanf("%d",&n);
    for (int i = 1;i <= n;i++) {
        scanf("%d%d",&p[i].x,&p[i].y);
        x[i] = p[i].x;
    }
    sort(x+1,x+1+n);
    int tot = unique(x+1,x+1+n) - x - 1;
    bit.init(tot);
    for (int i = 1;i <= n;i++) {
        p[i].x = lower_bound(x+1,x+tot+1,p[i].x) - x;
    }
    sort(p+1,p+1+n,cmpx);
    int cnt = 0;
    for (int i = 1;i < n;i++) {
        if (p[i].x == p[i+1].x) {
            line[++cnt] = {p[i].x,0,p[i].y,1};
            line[++cnt] = {p[i].x,0,p[i+1].y,-1};
        }
    }
    sort(p+1,p+1+n,cmpy);
    for (int i = 1;i < n;i++) {
        if (p[i].y == p[i+1].y) {
            line[++cnt] = {p[i].x,p[i+1].x,p[i].y,0};
        }
    }
    sort(line+1,line+1+cnt);
    int ans = 0;
    for (int i = 1;i <= cnt;i++) {
        if (!line[i].mark)
            ans += bit.qry(line[i].l,line[i].r);
        else
            bit.upd(line[i].l,line[i].mark);
    }
    printf("%d\n",ans + n);
    return 0;
}