HDU 4031 Attack（离线+线段树）

A - A

Time Limit:3000MS     Memory Limit:65768KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

Today is the 10th Annual of “September 11 attacks”, the Al Qaeda is about to attack American again. However, American is protected by a high wall this time, which can be treating as a segment with length N. Al Qaeda has a super weapon, every second it can attack a continuous range of the wall. American deployed N energy shield. Each one defends one unit length of the wall. However, after the shield defends one attack, it needs t seconds to cool down. If the shield defends an attack at kth second, it can’t defend any attack between (k+1)th second and (k+t-1)th second, inclusive. The shield will defend automatically when it is under attack if it is ready. 

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. 

 

Input

The beginning of the data is an integer T (T ≤ 20), the number of test case. 
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 

 

Output

For the ith case, output one line “Case i: ” at first. Then for each query, output one line containing one integer, the number of time the pth unit was successfully attacked when asked.

 

Sample Input

 23 7 2Attack 1 2Query 2Attack 2 3Query 2Attack 1 3Query 1Query 39 7 3Attack 5 5Attack 4 6Attack 3 7Attack 2 8Attack 1 9Query 5Query 3 

 

Sample Output

 Case 1:0101Case 2:32 

 

官方题解：线段树，标程是离线算法，按时间建立线段树，然后按区间的先后顺序插进去。

1001就是从1到n依次搞定每个点在那些时间被攻击过，由于每个攻击区间都是连续的，所以那些时间可以靠一个线段树来维护起 

————————————————————————————————————————————————————————————

其中一个正解是：离线处理，从1~n的墙按顺序处理答案。然后，可以注意到t最多只有50。

开一棵时间线段树，在我的代码中：

atk[x][i]代表在线段树结点x代表的时间区间中，防御壁一开始有 i 个单位时间的无法防御期，此时走过 x 中的时间，会受到的攻击次数。

empty[x][i]代表在线段树结点x代表的时间区间中，防御壁一开始有 i 个单位时间的无法防御期，此时走过 x 中的时间后，会有的无法防御时间。

 

PS：还有一种分块的在线做法：http://blog.renren.com/share/240115026/8571592218

#include <cstdio>#include <cstring>#include <algorithm>#include <cstring>#include <vector>using namespace std;const int MAXN = 20010;const int MAXT = MAXN << 2;const int MAXP = 55;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 * 2];int ans[MAXN];int T, n, q, t, ncnt, atime;void init() {    for(int i = 0; i <= n; ++i) qtime[i].clear(), qid[i].clear();    memset(ans, -1, q * sizeof(int));    ncnt = atime = 0;}#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 = 0; 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][0] = empty[x][0] = cnt[l] = 0;        for(int i = 1; i < t; ++i)            atk[x][i] = 0, empty[x][i] = i - 1;    } else {        build(ll, l, mid);        build(rr, mid + 1, r);        update(x);    }}void modify(int x, int l, int r, int a, int b) {    if(a <= l && r <= b) {        atk[x][0] = 0; empty[x][0] = cnt[a] ? t - 1 : 0;        for(int i = 1; i < t; ++i)            atk[x][i] = cnt[a], empty[x][i] = i - 1;    } else {        if(a <= mid) modify(ll, l, mid, a, b);        if(mid < b) modify(rr, mid + 1, r, a, b);        update(x);    }}void modify(int pos) {    modify(1, 1, 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 + 1, r, a, b, empty[ll][e]);        return res;    }}int query(int pos) {    if(pos == 0) return 0;    return query(1, 1, atime, 1, pos, 0);}char s[10];int main() {    scanf("%d", &T);    for(int kase = 1; kase <= T; ++kase) {        scanf("%d%d%d", &n, &q, &t);        init();        for(int i = 0, a, b; i < q; ++i) {            scanf("%s", s);            if(strcmp(s, "Attack") == 0) {                scanf("%d%d", &a, &b);                atime++;                attack[ncnt++] = Node(a, 1, atime);                attack[ncnt++] = Node(b + 1, -1, atime);            } else {                scanf("%d", &a);                qtime[a].push_back(atime);                qid[a].push_back(i);            }        }        sort(attack, attack + ncnt);        build(1, 1, atime);        int p = 0;        for(int i = 1; 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 = 0; k < qtime[i].size(); ++k)                ans[qid[i][k]] = query(qtime[i][k]);        }        printf("Case %d:\n", kase);        for(int i = 0; i < q; ++i)            if(ans[i] != -1) printf("%d\n", ans[i]);    }}