Educational Codeforces Round 4 D. The Union of k-Segments(★)

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题意：

给定n个区间，问你被覆盖至少k次的区间最少有多少个，并输出.

思路十分简洁，一个区间若被覆盖k次，则在这个区间前至少要出现k个线段的起点，且出现的起点减去终点数量不能小于k，这样排下序，线段起点+1，终点-1，判断下是否大于k一遍遍历下来就出答案了。

#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <cstdlib>#include <algorithm>#include <cmath>#include <vector>#include <set>#include <list>#include <queue>#include <map>#include <stack>using namespace std;#define L(i) i<<1#define R(i) i<<1|1#define INF  0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-3#define maxn 2000010#define MOD 100000007struct node{    int val,flag;    bool operator <(const node &a)const    {        return val < a.val || (val == a.val && flag < a.flag);    }}a[maxn];int n,k,ans[maxn];int main(){    int t,c = 1;    //scanf("%d",&t);    while(scanf("%d%d",&n,&k) != EOF)    {        for(int i = 0; i < n; i++)        {            int x,y;            scanf("%d%d",&x,&y);            a[2*i].val = x;            a[2*i].flag = 0;            a[2*i+1].val = y;            a[2*i+1].flag = 1;        }        sort(a,a+2*n);        int cnt = 0,num = 0;        for(int i = 0; i < 2*n; i++)        {            if(a[i].flag == 0)            {                cnt++;                if(cnt == k)                    ans[num++] = a[i].val;            }            else            {                if(cnt == k)                    ans[num++] = a[i].val;                cnt--;            }        }        printf("%d\n",num/2);        for(int i = 0; i < num; i+= 2)            printf("%d %d\n",ans[i],ans[i+1]);    }    return 0;}

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