CodeForces 754D Fedor and coupons (优先队列)

题意：给定n个优惠券，每张都有一定的优惠区间，然后要选k张，保证k张共同的优惠区间最大。

析：先把所有的优惠券按左端点排序，然后维护一个容量为k的优先队列，每次更新优先队列中的最小值，和当前的右端点，

之间的距离。优先队列只要存储右端点就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1LL << 60;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 3e5 + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
struct Node{
    int l, r, id;
    bool operator < (const Node &p) const{
        return l < p.l;
    }
};
Node a[maxn];

int main(){
    while(scanf("%d %d", &n, &m) == 2){
        for(int i = 0; i < n; ++i){
            scanf("%d %d", &a[i].l, &a[i].r);
            a[i].id = i + 1;
        }
        sort(a, a + n);
        priority_queue<int, vector<int>, greater<int> >pq;
        int ans = 0;
        int t = 0;
        for(int i = 0; i < n; ++i){
            pq.push(a[i].r);
            if(pq.size() > m)  pq.pop();
            int tmp = pq.top() - a[i].l + 1;
            if(pq.size() == m && ans < tmp){
                ans = tmp;
                t = a[i].l;
            }
        }

        printf("%d\n", ans);
        if(!ans){
            printf("1");
            for(int i = 2; i <= m; ++i)  printf(" %d", i);
            continue;
        }
        else{
            int cnt = 0;
            for(int i = 0; i < n && m; ++i) if(a[i].l <= t && a[i].r >= t + ans - 1){
                if(cnt)  putchar(‘ ‘);
                printf("%d", a[i].id);
                --m;
                ++cnt;
            }
        }
        printf("\n");
    }
    return 0;
}