Color the ball (segment tree)

Time limit:9000/3000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 11355 Accepted Submission (s): 5655

Problem descriptionn a row of balloons, numbered from left to right to 1,2,3....N. Each time a given 2 integers a B (a <= B), Lele will be riding his "little Pigeon" brand electric car from balloon a start to balloon B to each balloon painted once color. But n times after Lele has forgotten the first balloon has been painted several times, can you help him to figure out how many times each balloon has been painted color?

Input each test instance first behaves as an integer N, (n <= 100000). The next N rows, each line consists of 2 integers a B (1 <= a <= b <= N).
When n = 0, the input ends.

Output one row for each test instance, including n integers, and the number of numbers representing the number of times the balloon was painted in total.

Sample Input31 12 23 33 1 11 21 30

Sample OUTPUT1 1 13 2 1 Idea: Very typical segment tree, press test instructions to update the value of the interval

/*time memy 780ms 9132k by ORC 4 * * #include <cstdio> #include <cstring> #include <iostream>#    Include <algorithm>using namespace Std;const int N = 120005;int n,qans;struct node{int lt,rt,val; int add;}    S[4 * n];void Build (int v,int l,int r) {s[v].lt = l;    S[v].rt = R;    S[v].val = 0;    S[v].add = 0;    if (L = = r) return;    int m = (L + r) >> 1;    Build (V * 2, L, M); Build (V * 2 + 1, m + 1, R);} void update (int v,int l,int R) {if (s[v].lt = = L && s[v].rt = = r) {S[v].add + = 1; return;}        Only record increment, itself does not add, wait for the next query when first add if (s[v].add) {s[2 * v].add + = S[v].add;        S[2 * v + 1].add + = S[v].add;        S[v].val + = S[v].add;    S[v].add = 0;    } int m = (s[v].lt + s[v].rt) >> 1;    if (r <= m) update (2 * V, L, R);    else if (L > M) Update (2 * v + 1, L, R);        else {update (2 * V, L, M);    Update (2 * v + 1, m + 1, R); }}void query (int v,int l,int R) {if (s[v].add)//Query the time first look at the increment and pass   {s[2 * v].add + = S[v].add;        S[2 * v + 1].add + = S[v].add;    S[v].val + = S[v].add * (s[v].rt-s[v].lt + 1);//First add its own increment to its own s[v].add = 0;    } if (s[v].lt = = L && s[v].rt = = r) {Qans + = S[v].val; return;}    int m = (s[v].lt + s[v].rt) >> 1;    if (r <= m) query (2 * V, L, R);    else if (L > M) query (2 * v + 1, L, R);        else {query (2 * V, L, M);    Query (2 * v +1, M + 1, R);    }}void solve () {build (1, 1, N);    qans=0;    int A, B;        for (int i = 0;i < n; ++i) {cin>>a>>b;    Update (1, a, b);    } for (int i = 1;i < n; ++i) {qans=0;    Query (1, I, I);    cout<<qans<< ";    } qans=0;    Query (1, n, N); Cout<<qans<<endl;}    int main () {std::ios::sync_with_stdio (0);    while (Cin>>n && N) solve (); return 0;}

Color the ball (segment tree)