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;

}

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CodeForces 754D Fedor and coupons (优先队列)

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