zoj 3511 Cake Robbery(线段树)

problemCode=3511" target="_blank" style="">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;
}