HDU 4031 Attack(离线+线段树)(The 36th ACM/ICPC Asia Regional Chengdu Site —— Online Contest)
During the war, it is very important to understand the situation of both self and the enemy. So the commanders of American want to know how much time some part of the wall is successfully attacked. Successfully attacked means that the attack is not defended by the shield.
The first line of each test case is three integers, N, Q, t, the length of the wall, the number of attacks and queries, and the time each shield needs to cool down.
The next Q lines each describe one attack or one query. It may be one of the following formats
1. Attack si ti
Al Qaeda attack the wall from si to ti, inclusive. 1 ≤ si ≤ ti ≤ N
2. Query p
How many times the pth unit have been successfully attacked. 1 ≤ p ≤ N
The kth attack happened at the kth second. Queries don’t take time.
1 ≤ N, Q ≤ 20000
1 ≤ t ≤ 50
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std; const int MAXN = ;
const int MAXT = MAXN << ;
const int MAXP = ; struct Node {
int pos, op, time;
Node() {}
Node(int pos, int op, int time): pos(pos), op(op), time(time) {}
bool operator < (const Node &rhs) const {
return pos < rhs.pos;
}
}; vector<int> qtime[MAXN], qid[MAXN];
Node attack[MAXN * ];
int ans[MAXN];
int T, n, q, t, ncnt, atime; void init() {
for(int i = ; i <= n; ++i) qtime[i].clear(), qid[i].clear();
memset(ans, -, q * sizeof(int));
ncnt = atime = ;
} #define ll (x << 1)
#define rr (ll | 1)
#define mid ((l + r) >> 1)
int atk[MAXT][MAXP], empty[MAXT][MAXP];
int cnt[MAXN]; void update(int x) {
for(int i = ; i < t; ++i) {
int t = empty[ll][i];
atk[x][i] = atk[ll][i] + atk[rr][t];
empty[x][i] = empty[rr][t];
}
} void build(int x, int l, int r) {
if(l == r) {
atk[x][] = empty[x][] = cnt[l] = ;
for(int i = ; i < t; ++i)
atk[x][i] = , empty[x][i] = i - ;
} else {
build(ll, l, mid);
build(rr, mid + , r);
update(x);
}
} void modify(int x, int l, int r, int a, int b) {
if(a <= l && r <= b) {
atk[x][] = ; empty[x][] = cnt[a] ? t - : ;
for(int i = ; i < t; ++i)
atk[x][i] = cnt[a], empty[x][i] = i - ;
} else {
if(a <= mid) modify(ll, l, mid, a, b);
if(mid < b) modify(rr, mid + , r, a, b);
update(x);
}
} void modify(int pos) {
modify(, , atime, pos, pos);
} int query(int x, int l, int r, int a, int b, int e) {
if(a <= l && r <= b) {
return atk[x][e];
} else {
int res = query(ll, l, mid, a, b, e);
if(mid < b) res += query(rr, mid + , r, a, b, empty[ll][e]);
return res;
}
} int query(int pos) {
if(pos == ) return ;
return query(, , atime, , pos, );
} char s[]; int main() {
scanf("%d", &T);
for(int kase = ; kase <= T; ++kase) {
scanf("%d%d%d", &n, &q, &t);
init();
for(int i = , a, b; i < q; ++i) {
scanf("%s", s);
if(strcmp(s, "Attack") == ) {
scanf("%d%d", &a, &b);
atime++;
attack[ncnt++] = Node(a, , atime);
attack[ncnt++] = Node(b + , -, atime);
} else {
scanf("%d", &a);
qtime[a].push_back(atime);
qid[a].push_back(i);
}
}
sort(attack, attack + ncnt); build(, , atime);
int p = ;
for(int i = ; i <= n; ++i) {
while(p < ncnt && attack[p].pos == i) {
cnt[attack[p].time] += attack[p].op;
modify(attack[p++].time);
}
for(size_t k = ; k < qtime[i].size(); ++k)
ans[qid[i][k]] = query(qtime[i][k]);
} printf("Case %d:\n", kase);
for(int i = ; i < q; ++i)
if(ans[i] != -) printf("%d\n", ans[i]);
}
}
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