线段树

1、Codeforces 377D Developing Game

解题思路：

参考：http://codeforces.com/blog/entry/10157

http://www.cnblogs.com/qscqesze/p/5541837.html

转化为扫描线问题

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <queue>#include <vector>#include <stack>#include <map>#include <set>#include <cmath>#include <cctype>#include <ctime>#include <cassert>using namespace std;#define REP(i, n) for (int i = 0; i < (n); ++i)#define eps 1e-6#define PI acos(-1.0)#define lc id << 1#define rc id << 1 | 1#define lson low, mid, lc#define rson mid + 1, high, rctypedef long long ll;typedef pair<int, int> pii;struct Node {    int x, y1, y2, flag;    bool operator < (const Node& a) const {        if (x == a.x) { return flag > a.flag; } return x < a.x;    }};const int INF = 0x7fffffff;const int maxn = 6e5 + 10;vector<int> y;int n, id1, id2, Size, ans = 0, len_t, ans_l, ans_r, cnt = 0;Node node[maxn];int Left[maxn], v[maxn], Right[maxn], max_cnt[maxn * 4], max_r[maxn * 4], lazy[maxn * 4];int id(int x) { return lower_bound(y.begin(), y.end(), x) - y.begin() + 1; }void build(int low, int high, int id) {    max_r[id] = y[high - 1]; if (low == high) { return; }    int mid = (low + high) / 2; build(lson); build(rson);}inline void push_up(int id) {    if (max_cnt[rc] >= max_cnt[lc]) { max_r[id] = max_r[rc]; } else { max_r[id] = max_r[lc]; }    max_cnt[id] = max(max_cnt[lc], max_cnt[rc]);}inline void push_down(int id) {    if (lazy[id] == 0) { return; } max_cnt[lc] += lazy[id]; lazy[lc] += lazy[id];    max_cnt[rc] += lazy[id]; lazy[rc] += lazy[id]; lazy[id] = 0;}void update(int l, int r, int key, int low, int high, int id) {    if (l == low && r == high) { max_cnt[id] += key; lazy[id] += key; return; }    int mid = (low + high) / 2; push_down(id);    if (r <= mid) { update(l, r, key, lson); }    else if (l >= mid + 1) { update(l, r, key, rson); }    else { update(l, mid, key, lson); update(mid + 1, r, key, rson); }    push_up(id);}int main() {#ifdef __AiR_H    freopen("in.txt", "r", stdin);//    freopen("out.txt", "w", stdout);#endif // __AiR_H    scanf("%d", &n);    REP(i, n) {        scanf("%d %d %d", &Left[i], &v[i], &Right[i]); y.push_back(v[i]); y.push_back(Right[i]);        node[i * 2] = Node{Left[i], v[i], Right[i], 1}; node[i * 2 + 1] = Node{v[i], v[i], Right[i], -1};    }    sort(y.begin(), y.end()); y.erase(unique(y.begin(), y.end()), y.end());    n *= 2; sort(node, node + n); Size = y.size(); build(1, Size, 1);    REP(i, n) {        id1 = id(node[i].y1); id2 = id(node[i].y2);        update(id1, id2, node[i].flag, 1, Size, 1);        if (max_cnt[1] > ans) { ans = max_cnt[1]; ans_l = node[i].x; ans_r = max_r[1]; }    }    printf("%d\n", ans);    REP(i, n) {        if (Left[i] <= ans_l && ans_r <= Right[i] && ans_l <= v[i] && v[i] <= ans_r) {            printf(++cnt == 1 ? "%d" : " %d", i + 1);        }    }    printf("\n");#ifdef __AiR_H    printf("Time used = %.2fs\n", (double)clock() / CLOCKS_PER_SEC);#endif // __AiR_H    return 0;}

2、HDU 5722 Jewelry

参考：http://blog.csdn.net/jtjy568805874/article/details/51944740

PS：矩形变了样子，然后就不会求矩形并的面积了QwQ。。。

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <queue>#include <vector>#include <stack>#include <map>#include <set>#include <cmath>#include <cctype>#include <ctime>#include <cassert>using namespace std;#define REP(i, n) for (int i = 0; i < (n); ++i)#define eps 1e-6#define PI acos(-1.0)#define lc id << 1#define rc id << 1 | 1#define lson low, mid, lc#define rson mid + 1, high, rctypedef long long ll;typedef pair<int, int> pii;struct Node {    int x, y1, y2, flag;    bool operator < (const Node& a) const {        if (x == a.x) { return flag > a.flag; } return x < a.x;    }};const int INF = 0x7fffffff;const int maxn = 2e5 + 10;int T, n, x, A, N = 0;ll ans = 0;Node node[maxn];int cnt[maxn * 4], len[maxn * 4];vector<pii> v;void build(int low, int high, int id) {    cnt[id] = len[id] = 0; if (low == high) { return; }    int mid = (low + high) / 2; build(lson); build(rson);}inline void push_up(int id, int low, int high) {    if (cnt[id]) { len[id] = high - low + 1; return; }    len[id] = len[lc] + len[rc];}void update(int l, int r, int key, int low, int high, int id) {    if (l == low && r == high) {        cnt[id] += key; if (low == high && cnt[id] == 0) { len[id] = 0; return; }        push_up(id, low, high); return;    }    int mid = (low + high) / 2;    if (r <= mid) { update(l, r, key, lson); }    else if (l >= mid + 1) { update(l, r, key, rson); }    else { update(l, mid, key, lson); update(mid + 1, r, key, rson); }    push_up(id, low, high);}int main() {#ifdef __AiR_H    freopen("in.txt", "r", stdin);//    freopen("out.txt", "w", stdout);#endif // __AiR_H    scanf("%d", &T);    while (T--) {        v.clear(); N = 0;        scanf("%d %d", &n, &x); REP(i, n) { scanf("%d", &A); v.push_back(make_pair(A, i)); }        sort(v.begin(), v.end()); int l1, r1, l2, r2;        REP(i, n) {            if (i - x + 1 < 0 || v[i].first != v[i - x + 1].first) { continue; }            l1 = 0; r1 = v[i - x + 1].second; l2 = v[i].second; r2 = n - 1;            if (i - x >= 0 && v[i - x].first == v[i].first) { l1 = v[i - x].second + 1; }            if (i + 1 < n && v[i + 1].first == v[i].first) { r2 = v[i + 1].second - 1; }            ++l1; ++r1; ++l2; ++r2;            node[N++] = Node{l1, l2, r2, 1}; node[N++] = Node{r1, l2, r2, -1};        }        sort(node, node + N);  build(1, n, 1); ans = 0;        for (int i = 0, j; i < N; i = j) {            if (i > 0) { ans += (ll)(node[i].x - node[i - 1].x - 1) * len[1]; }            j = i;            while (j < N && node[j].x == node[i].x && node[j].flag == 1) {                update(node[j].y1, node[j].y2, node[j].flag, 1, n, 1); ++j;            }            ans += len[1];            while (j < N && node[j].x == node[i].x && node[j].flag == -1) {                update(node[j].y1, node[j].y2, node[j].flag, 1, n, 1); ++j;            }        }        printf("%I64d\n", ans);    }#ifdef __AiR_H    printf("Time used = %.2fs\n", (double)clock() / CLOCKS_PER_SEC);#endif // __AiR_H    return 0;}