题意: 贴海报 后贴的会覆盖前面 求从Front能看到几张海报
分析: 区间离散化之后 就是一个裸的区间染色啦 套用线段树区间更新就可以搞啦
代码:
// // Created by TaoSama on 2015-09-16 // Copyright (c) 2015 TaoSama. All rights reserved. // //#pragma comment(linker, "/STACK:1024000000,1024000000") #include <algorithm> #include <cctype> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <string> #include <set> #include <vector> using namespace std; #define pr(x) cout << #x << " = " << x << " " #define prln(x) cout << #x << " = " << x << endl const int N = 1e4 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7; int n, ans; int row[6 * N * 4]; int x1[N], x2[N]; bool vis[N]; #define root 1, n, 1 #define lson l, m, rt << 1 #define rson m + 1, r, rt << 1 | 1 void push_down(int rt) { if(row[rt]) { row[rt << 1] = row[rt << 1 | 1] = row[rt]; row[rt] = 0; } } void update(int L, int R, int c, int l, int r, int rt) { if(L <= l && r <= R) { row[rt] = c; return; } int m = l + r >> 1; push_down(rt); if(L <= m) update(L, R, c, lson); if(R > m) update(L, R, c, rson); } void query(int l, int r, int rt) { if(row[rt]) { if(!vis[row[rt]]) ++ans; vis[row[rt]] = true; return; } if(l == r) return; int m = l + r >> 1; query(lson); query(rson); } int compress(int *x1, int *x2, int w) { vector<int> xs; for(int i = 1; i <= n; ++i) { for(int d = -1; d <= 1; ++d) { int tx1 = x1[i] + d, tx2 = x2[i] + d; if(tx1 >= 1 && tx1 <= w) xs.push_back(tx1); if(tx2 >= 1 && tx2 <= w) xs.push_back(tx2); } } sort(xs.begin(), xs.end()); xs.resize(unique(xs.begin(), xs.end()) - xs.begin()); for(int i = 1; i <= n; ++i) { x1[i] = lower_bound(xs.begin(), xs.end(), x1[i]) - xs.begin() + 1; x2[i] = lower_bound(xs.begin(), xs.end(), x2[i]) - xs.begin() + 1; } return xs.size(); } int main() { #ifdef LOCAL freopen("in.txt", "r", stdin); // freopen("out.txt","w",stdout); #endif ios_base::sync_with_stdio(0); int t; scanf("%d", &t); while(t--) { scanf("%d", &n); for(int i = 1; i <= n; ++i) scanf("%d%d", x1 + i, x2 + i); int w = compress(x1, x2, 1e7); memset(row, 0, sizeof row); memset(vis, false, sizeof vis); for(int i = 1; i <= n; ++i) update(x1[i], x2[i], i, 1, w, 1); ans = 0; query(1, w, 1); printf("%d\n", ans); } return 0; }