LightOJ - 1301 Monitoring Processes(树状数组)

来源：互联网发布：淘宝新政策编辑：程序博客网时间：2024/05/20 04:50

题目大意：给你N个区间，问区间覆盖的最多次数

解题路上：坐标比较大，得离散话，然后树状数组维护每个点被覆盖了几次即可，最后再取最大值

#include <cstdio>#include <cstring>#include <map>#include <algorithm>using namespace std;const int N = 50010;struct Node {    int l, r;    Node() {}    Node(int l, int r): l(l), r(r) {}}node[N];int bit[2 * N], num[N * 2];int n, cnt, cas = 1;map<int,int> Map;void init() {    scanf("%d", &n);    Map.clear();    cnt = 0;    for (int i = 0; i < n; i++) {        scanf("%d%d", &node[i].l, &node[i].r);        if (!Map[node[i].l]) {            Map[node[i].l] = ++cnt;            num[cnt] = node[i].l;        }        if (!Map[node[i].r])  {            Map[node[i].r] = ++cnt;            num[cnt] = node[i].r;        }    }    sort(num + 1, num + 1 + cnt);}int find(int pos) {    int l = 1, r = cnt;    while (l <= r) {        int mid = (l + r) >> 1;        if (num[mid] == pos) return mid;        else if (num[mid] > pos) r = mid - 1;        else l = mid + 1;    }    return -1;}inline int lowbit(int x) {    return x & (-x);}void Modify(int x, int val) {    while (x <= cnt) {        bit[x] += val;        x += lowbit(x);    }}int Query(int x) {    int ans = 0;    while (x) {        ans += bit[x];        x -= lowbit(x);    }    return ans;}void solve() {    memset(bit, 0, sizeof(bit));    for (int i = 0; i < n; i++) {        int l = find(node[i].l);        int r = find(node[i].r) + 1;        Modify(l, 1);        Modify(r, -1);    }    int ans = 0;    for (int i = 1; i <= cnt; i++) {        ans = max(ans, Query(i));    }    printf("Case %d: %d\n", cas++, ans);}int main() {    int test;    scanf("%d", &test);    while (test--) {        init();        solve();    }    return 0;}

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