11402-ahoy, pirates! (segment Tree interval update (labeling overlapping processing))

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Test instructions: There are 3 kinds of interval operations, all of which turn a range into 1, a range of 0, a range of 1 into 0, 0 to 1.

Idea: The first two operations is the most basic interval update, using the lazy tag, however, the 3rd operation is a bit of trouble, if only update the current node corresponding to the large interval, then it contains a small interval is updated when the error occurs, the cause of the error is because of the overlap and collision of the tag. Obviously, this is a typical problem, dealing with the problem of tag collisions.

The core of the problem is resolving collisions so that each point has at most one marker at a time.

So what do we do? We can take a look at the equivalent of what happens when two markers collide:

When we want to update a node o:

If we are going to update the interval, we will overwrite it without having to control the original tag. But if the interval is reversed:

If the node already has a Mark V

So:

When v = = 1, indicating that the interval was once all changed to 1 of the operation, equivalent to this time all become 0, so the lazy mark eventually becomes 0.

When v = = 0, in the same vein, the lazy sign eventually becomes 1.

When the V = = 2, indicating that the interval has a full reversal of the operation has not been implemented, equivalent to this time not reversed, so the lazy mark eventually becomes-1

When v = =-1, indicating that there is no operation on the interval, equivalent to the need for reversal, the lazy mark eventually 2

In fact, it is an equivalent conversion, when more collisions will involve a processing of the shun-seeking problem.

Also, if you update the SUM[V when updating node V, then it involves the update of sum[v] in the equivalent conversion, so don't think wrong.

See the code for details:

#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include < string> #include <vector> #include <stack> #include <bitset> #include <cstdlib> #include < cmath> #include <set> #include <list> #include <deque> #include <map> #include <queue># Define MAX (a) > (b)? ( A):(B) #define MIN (a) < (b) ( A):(B)) using namespace Std;typedef long Long ll;const double PI = ACOs ( -1.0); const double EPS = 1e-6;const int mod = 10000 00000 + 7;const int INF = 1000000000;const int MAXN = 1024010;int t,n,m,q,l,r,kase=0,sum[maxn<<2],setv[maxn<< 2],a[maxn],cnt = 0;void pushup (int o) {Sum[o] = sum[o<<1] + sum[o<<1|1];}        void pushdown (int l, int r, int o) {if (Setv[o]! =-1) {int m = (L + r) >> 1;            if (setv[o] = = 1) {sum[o<<1] = (m-l + 1);            SUM[O&LT;&LT;1|1] = (r-m);        SETV[O&LT;&LT;1] = setv[o<<1|1] = Setv[o];     }   else if (setv[o] = = 0) {sum[o<<1] = 0;            Sum[o<<1|1] = 0;        SETV[O&LT;&LT;1] = setv[o<<1|1] = Setv[o];                } else {if (setv[o<<1] = = 1) {setv[o<<1] = 0;            Sum[o<<1] = 0;                } else if (setv[o<<1] = = 0) {setv[o<<1] = 1;            SUM[O&LT;&LT;1] = (m-l + 1);            } else if (setv[o<<1] = = 2) setv[o<<1] = 1, sum[o<<1] = m-l + 1-sum[o<<1];                else {setv[o<<1] = 2;            SUM[O&LT;&LT;1] = m-l + 1-sum[o<<1];                } if (setv[o<<1|1] = = 1) {setv[o<<1|1] = 0;            Sum[o<<1|1] = 0;                } else if (setv[o<<1|1] = = 0) {setv[o<<1|1] = 1;            SUM[O&LT;&LT;1|1] = (r-m); } else if (setv[o<<1|1] = = 2)SETV[O&LT;&LT;1|1] = 1, sum[o<<1|1] = r-m-sum[o<<1|1];                else {setv[o<<1|1] = 2;            SUM[O&LT;&LT;1|1] = r-m-sum[o<<1|1];    }} Setv[o] =-1;    }}void Build (int l, int r, int o) {int m = (L + r) >> 1;    Setv[o] =-1;        if (L = = r) {++cnt;        if (a[cnt]) sum[o] = 1;        else Sum[o] = 0;    return;    } build (L, M, o<<1);    Build (M+1, R, O<<1|1); Pushup (o);}    void Update (int l, int r, int v, int l, int r, int o) {int m = (L + R) >> 1;            if (l <= l && R <= R) {if (v = = 1) {Sum[o] = (r-l + 1);        Setv[o] = 1;            } else if (v = = 0) {Sum[o] = 0;        Setv[o] = 0;                } else {if (setv[o] = = 1) {Setv[o] = 0;            Sum[o] = 0;                } else if (setv[o] = = 0) {Setv[o] = 1; Sum[o] = r-l + 1;                } else if (setv[o] = = 1) {Setv[o] = 2;            Sum[o] = r-l + 1-sum[o];        } else Setv[o] =-1, Sum[o] = r-l + 1-sum[o];    } return;    } pushdown (L, R, O);    if (l <= m) update (L, R, V, L, M, o<<1);    if (M < R) Update (L, R, V, M+1, R, O<<1|1); Pushup (o);}    int query (int l, int r, int l, int r, int o) {int m = (L + R) >> 1;    if (l <= l && R <= R) {return sum[o];    } pushdown (L, R, O);    int ans = 0;    if (l <= m) ans + = query (l, R, L, M, o<<1);    if (M < r) ans + = query (L, R, M+1, R, O<<1|1);    Pushup (o); return ans;}    Char S[100];int main () {scanf ("%d", &t);        while (t--) {scanf ("%d", &m);        int n = 0;            for (int i=0;i<m;i++) {scanf ("%d%s", &cnt,s);            int len = strlen (s); while (cnt--) {for (int j=0;j<len;j++) {A[++n] = s[j]-' 0 ';        }}} cnt = 0;        Build (1, N, 1);        scanf ("%d", &q);        printf ("Case%d:\n", ++kase);        int cc = 0;            while (q--) {scanf ("%s%d%d", S,&l,&r); l++;            r++;            if (s[0] = = ' F ') update (L, R, 1, 1, N, 1);            else if (s[0] = = ' E ') update (L, r, 0, 1, N, 1);            else if (s[0] = = ' I ') update (L, R, 2, 1, N, 1);        else printf ("q%d:%d\n", ++CC, Query (L, R, 1, N, 1)); }} return 0;}

11402-ahoy, pirates! (segment Tree interval update (labeling overlapping processing))