HDU 5367 digger

Problem Description

AFa have n mountains. These mountains in a line。All of mountains have same height initially. Every day , AFa would request ZJiaQ to cut some mountains or put stones in some mountains. And request ZJiaQ report how many mountains belong to “high mountain line”
when all of mountains in a range have same height, and higher than the nearest mountain in left and the nearest mountain in right. the range of mountains called “high mountain line”
of course ,the mountain in the most left and most right can’t be one of “high mountain line”

 

Input

  There are multiply case
In each case, the first line contains 3 integers: n(1<=n<=10^9) , q ( 1 <= q <= 50000), r(0 <= r<= 1000 ).n is the number of mountains. q is the number of days. r is the initial height .
In the next q lines, each line contains 3 integers: l, r, val(1<=l<=r<=n, -1000 <= val <= 1000). Means in the day the mountains’ height in range[l,r] have added by val. but you should let the l,r,val xor ans to get true l,r,val. ans is the answer you printed. 

 

Output

Print q lines, each line contains a single answer.

 

Sample Input

5 5 04 5 872 5 -483 3 174 5 -1715 5 -494

 

Sample Output

00011
线段树动态建点
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio>#include<algorithm>using namespace std;const int maxn = 1200005;int n, m, h, l, r;struct node{int lh, rh, lr, rl, lf, rf,l, r;void clear(){ l = r = lh = rh = lr = rl = lf = rf = 0; }};struct ST{int L[maxn], R[maxn], f[maxn], sum[maxn];node u[maxn];int n, tt;int newnode(int l, int r){f[++tt] = 0;L[tt] = R[tt] = sum[tt] = 0;u[tt].clear();u[tt].l = u[tt].rl = l;u[tt].r = u[tt].lr = r;return tt;}void build(int x){ n = x; tt = 1; sum[1] = L[1] = R[1] = f[1] = 0;u[1].clear();u[0].clear();u[1].rl = u[1].l = 1;u[1].lr = u[1].r = n;}void update(int x, int l, int r, int lx, int rx){int llh = u[lx].lh + sum[lx];int lrh = u[lx].rh + sum[lx];int rlh = u[rx].lh + sum[rx];int rrh = u[rx].rh + sum[rx];int mid = (l + r) >> 1;u[x].lh = llh;u[x].rh = rrh;if (u[lx].lr == u[lx].r && llh == rlh){u[x].lr = rx ? u[rx].lr : r;u[x].lf = u[rx].lf;}else{u[x].lr = lx ? u[lx].lr : mid;if (u[lx].lr == u[lx].r) u[x].lf = llh > rlh ? 1 : 0;else u[x].lf = u[lx].lf;}if (u[rx].rl == u[rx].l&&rrh == lrh){u[x].rl = lx ? u[lx].rl : l;u[x].rf = u[lx].rf;}else{u[x].rl = rx ? u[rx].rl : (mid + 1);if (u[rx].rl == u[rx].l) u[x].rf = rrh > lrh ? 1 : 0;else u[x].rf = u[rx].rf;}f[x] = f[lx] + f[rx];if (u[lx].rl > u[lx].l&&lrh > rlh&&u[lx].rf) f[x] += u[lx].r - u[lx].rl + 1;if (u[rx].lr < u[rx].r&&rlh > lrh&&u[rx].lf) f[x] += u[rx].lr - u[rx].l + 1;if (u[lx].rl > u[lx].l&&u[rx].lr < u[rx].r&&lrh == rlh&&u[lx].rf&&u[rx].lf)f[x] += u[rx].lr - u[lx].rl + 1;}void add(int x, int l, int r, int ll, int rr, int v){if (ll <= l&&r <= rr) sum[x] += v; else{int mid = (l + r) >> 1;if (ll <= mid){if (!L[x]) L[x] = newnode(l, mid);add(L[x], l, mid, ll, rr, v);}if (rr > mid){if (!R[x]) R[x] = newnode(mid + 1, r);add(R[x], mid + 1, r, ll, rr, v);}update(x, l, r, L[x], R[x]);}}}st;int main(){while (scanf("%d%d%d", &n, &m, &h) != EOF){st.build(n);while (m--){scanf("%d%d%d", &l, &r, &h);l ^= st.f[1];r ^= st.f[1];h ^= st.f[1];if (l > r) swap(l, r);st.add(1, 1, n, l, r, h);printf("%d\n", st.f[1]);}} return 0;}