Segment tree + discretized POJ 2528 Mayor ' s posters

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Test instructions: Put a poster on one wall, in order, and ask how many different posters are in the end (those that are not covered or just partially covered)

Analysis: This data range is very large, directly engaged in time-out + hyper-memory, need discretization: discretization is simply to take the values we need to use, such as interval [1000,2000],[1990,2012] We do not use [-∞,999][1001,1989][ 1991,1999][2001,2011][2013,+∞] These values, so I just need to 1000,1990,2000,2012 is enough, to map it to 0,1,2,3, the complexity is greatly lowered, so discretization to save all the required values, Once sorted, they are mapped to 1~n, which makes the complexity much smaller. The difficulty of this problem is that each number actually represents a unit length (not a point), so that ordinary discretization will cause many errors (including my previous code, POJ the data is very weak). The following two simple examples should be presented to illustrate the drawbacks of common discretization:

Example one: 1-10 1-4 5-10
Example two: 1-10 1-4 6-10
Common discretization has become a [1,4][1,2][3,4]

Match (example one):
To solve this flaw, we can add some processing to the sorted array, for example [1,2,6,10]
If the adjacent digit spacing is greater than 1, add any number to it, such as add [1,2,3,6,7,10], and then do the line segment tree. --copy from Notonlysuccess
Harvesting: discretization Techniques

Code:

#include <cstdio> #include <cstring> #include <algorithm>using namespace std; #define Lson L, Mid, RT << 1#define Rson mid + 1, R, RT << 1 |    1const int N = 1e4 + 10;const int INF = 0x3f3f3f3f;int ans;struct ST {int col[n<<4];    BOOL Vis[n];        void init (void) {memset (col,-1, sizeof (COL));    Memset (Vis, false, sizeof (VIS));            } void Push_down (int rt) {if (Col[rt]! =-1) {col[rt<<1] = col[rt<<1|1] = Col[rt];        COL[RT] =-1;            }} void Updata (int ql, int qr, int c, int l, int r, int rt) {if (QL <= l && r <= qr) {    COL[RT] = c;        return;        } push_down (RT);        int mid = (L + r) >> 1;        if (QL <= mid) Updata (QL, QR, C, Lson);    if (QR > Mid) updata (QL, QR, C, Rson); } void query (int l, int r, int rt) {if (Col[rt]! =-1) {if (!vis[col[rt]]) {ans  ++; vis[COL[RT]] = true;        } return;        } if (L = = r) return;        int mid = (L + r) >> 1;        Query (Lson);    Query (Rson);   }}st;int L[n], r[n];int x[n<<2];int main () {int T, N;        scanf ("%d", &t), while (T--) {scanf ("%d", &n);        int tot = 0;            for (int i=0; i<n; ++i) {scanf ("%d%d", &l[i], &r[i]);            x[tot++] = L[i];        x[tot++] = R[i];        } sort (X, x+tot); int k = 1;        for (int i=1; i<tot; ++i) {if (X[i]! = X[i-1]) x[k++] = X[i];        } for (int i=k-1; i>=1;-i) {if (X[i]! = X[i-1] + 1) x[k++] = x[i-1] + 1;        }sort (X, x+k);        St.init ();            for (int i=0; i<n; ++i) {int QL = Lower_bound (x, X+k, L[i])-X;            int qr = Lower_bound (x, X+k, R[i])-X;        St.updata (QL, QR, I, 0, K, 1); }ans = 0;st.query (0, K, 1);p rintf ("%d\n", ans); return 0;}

　　

Segment tree + discretized POJ 2528 Mayor ' s posters