浙大17年校赛(ZOJ 3953) Intervals[贪心]

题意:给了n个区间，要求你删去最少的区间，使任意三个区间 a,b,c 不存在 a与b相交，b与c相交，c与a相交 的情况。

分析：比赛时候看到这题，还以为要用什么高深的数据结构，出来听他们说贪心后马上就想到了思路，，还是太菜了。。首先，我们先将n个区间按x从小到大，再y从小到大排序，然后遍历，如果三个区间满足 上述的情况，那么我们将y最大的删去，y相等将x最小的删去。过程中维护两个区间tmp1,tmp2,表示前两个相交的区间是什么，然后就是一些分类，具体看代码。

以下是代码。

#include<iostream>#include<cstdio>#include<algorithm>#include<queue>#include<map>#include<set>#include<stack>#include<cstring>#include<string>#include<vector>#include<iomanip>#include<unordered_set>#include<unordered_map>#include<cmath>#include<list>#include<bitset>using namespace std;#define ull unsigned long long#define ll long long#define lson l,mid,id<<1#define rson mid+1,r,id<<1|1typedef pair<int, int>pii;typedef pair<ll, ll>pll;typedef pair<double, double>pdd;const double eps = 1e-6;const int MAXN = 100005;const int MAXM = 5005;const ll LINF = 0x3f3f3f3f3f3f3f3f;const int INF = 0x3f3f3f3f;const double FINF = 1e18;const ll MOD = 1000000007;struct lx {lx(){}lx(int _x, int _y, int _id){x = _x, y = _y, id = _id;}int x, y, id;bool operator<(const lx &a)const {if (x == a.x)return y < a.y;else return x < a.x;}}tmp[50005];lx getid(lx a, lx b, lx c){int id, x, y;if (a.y > b.y){y = a.y;x = a.x;id = a.id;}else if (a.y == b.y){if (a.x < b.x){y = a.y; x = a.x;id = a.id;}else{y = b.y; x = b.x;id = b.id;}}else{y = b.y; x = b.x; id = b.id;}if (c.y > y)x = c.x, y = c.y, id = c.id;else if (c.y == y){if (c.x < x){x = c.x, y = c.y, id = c.id;}}return lx(x, y, id);}int main(){int t, n, cnt, num;scanf("%d", &t);lx tmp1, tmp2;vector<int>ans;while (t--){num = 0;cnt = 0;ans.clear();scanf("%d", &n);for (int i = 1; i <= n; ++i){scanf("%d%d", &tmp[i].x, &tmp[i].y);tmp[i].id = i;}sort(tmp + 1, tmp + n + 1);//for (int i = 1; i <= n; ++i)cout << tmp[i].x << " " << tmp[i].y << " " << tmp[i].id << endl;if (tmp[2].x >= tmp[1].x && tmp[2].x <= tmp[1].y){tmp1 = tmp[1], tmp2 = tmp[2];num = 2;}else{tmp1 = tmp[2], num = 1;}for (int i = 3; i <= n; ++i){if (num == 2){if (tmp[i].x >= tmp2.x && tmp[i].x <= tmp2.y){if (tmp[i].x >= tmp1.x && tmp[i].x <= tmp1.y){cnt++;lx uu = getid(tmp[i], tmp1, tmp2);ans.push_back(uu.id);if (uu.id == tmp1.id){tmp1 = tmp2; tmp2 = tmp[i];}else if (uu.id == tmp2.id)tmp2 = tmp[i];}else{tmp1 = tmp2;tmp2 = tmp[i];}}else{if (tmp[i].x >= tmp1.x && tmp[i].x <= tmp1.y){tmp2 = tmp[i];}else{tmp1 = tmp[i]; num = 1;}}}else{if (tmp[i].x >= tmp1.x && tmp[i].x <= tmp1.y){tmp2 = tmp[i]; num = 2;}else{tmp1 = tmp[i];}}}sort(ans.begin(), ans.end());printf("%d\n", cnt);if (cnt == 0)printf("\n");for (int i = 0; i < ans.size(); ++i){if (i == ans.size() - 1)printf("%d\n", ans[i]);else printf("%d ", ans[i]);}}}