zoj 3511 Cake Robbery(线段树)

题目链接：zoj 3511 Cake Robbery

题目大意：就是有一个N边形的蛋糕。切M刀，从中挑选一块边数最多的。保证没有两条边重叠。

解题思路：有多少个顶点即为有多少条边，所以直接依照切刀切掉点的个数排序，然后用线段树维护剩下的还有哪些点。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>using namespace std;const int maxn = 10005;#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], s[maxn << 2];inline void pushdown(int u) {if (s[u] == 0)s[lson(u)] = s[rson(u)] = 0;
}inline void pushup(int u) {s[u] = s[lson(u)] + s[rson(u)];
}void build (int u, int l, int r) {lc[u] = l;rc[u] = r;if (l == r) {s[u] = 1;return;}int mid = (l + r) / 2;build(lson(u), l, mid);build(rson(u), mid + 1, r);pushup(u);
}void modify (int u, int l, int r) {if (l > r)return;if (l <= lc[u] && rc[u] <= r) {s[u] = 0;return;}pushdown(u);int mid = (lc[u] + rc[u]) / 2;if (l <= mid)modify(lson(u), l, r);if (r > mid)modify(rson(u), l, r);pushup(u);
}int N, M;
struct Seg {int l, r, c;Seg (int l = 0, int r = 0) {this->l = l;this->r = r;this->c = r - l + 1;}friend bool operator < (const Seg& a, const Seg& b) {return a.c < b.c;}
};
vector<Seg> vec;int main () {while (scanf("%d%d", &N, &M) == 2) {int l, r, ans = 0;build(1, 1, N);vec.clear();while (M--) {scanf("%d%d", &l, &r);if (l > r) swap(l, r);vec.push_back(Seg(l, r));}sort(vec.begin(), vec.end());for (int i = 0; i < vec.size(); i++) {int tmp = s[1];modify(1, vec[i].l + 1, vec[i].r - 1);ans = max(ans, tmp - s[1] + 2);}printf("%d\n", max(ans, s[1]));}return 0;
}