HDOJ 5214 Movie-优快云博客

本文探讨了在电影制作中如何通过大模拟和贪心算法来解决多个重要场景时间调度问题，确保在有限时间内能观赏到至少三个不同的关键场景。详细介绍了算法思路、实现过程以及注意事项。

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题意：电影中有n个重要的场景，每个场景出现的时间在[l,r]之间，求在这部电影中是否能看到三个不同的重要场景

链接：http://acm.hdu.edu.cn/showproblem.php?pid=5214

思路：大模拟，按照题中所给的公式求出所有的时间区间后贪心

注意点：“After all the intervals are generated, swap the i-th interval's Li and Ri if Li > Ri ”在求出所有序列后再对不满足条件的l,r进行交换。直接利用三次循环进行贪心，排序后贪心会超时。

以下为AC代码：

Run ID	Submit Time	Judge Status	Pro.ID	Exe.Time	Exe.Memory	Code Len.	Language	Author
13617864	2015-05-03 21:15:31	Accepted	5214	904MS	79856K	3892 B	G++	luminous11

/* 
***********************************************
*# @Author  : Luminous11 (573728051@qq.com)
*# @Date    : 2015-05-02 12:44:01
*# @Link    : http://blog.youkuaiyun.com/luminous11
*********************************************** 
*/

#include <algorithm>
#include <iostream>
#include <climits>
#include <iomanip>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <deque>
#include <list>
#include <map>
#include <set>
//#include <unordered_map>
//#include <unordered_set>
#define pb push_back
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define READ(f) freopen(f, "r", stdin)
#define WRITE(f) freopen(f, "w", stdout)
#define clr(a, v) memset( a , v , sizeof(a) )
#define RS(s) scanf ( "%s", s )
#define RDI(a) scanf ( "%d", &a )
#define RDII(a, b) scanf ( "%d%d", &a, &b )
#define RDIII(a, b, c) scanf ( "%d%d%d", &a, &b, &c )
#define PL() printf ( "\n" )
#define PI(a) printf ( "%d", a )
#define PIL(a) printf ( "%d\n", a )
#define PSL(s) printf ( "%s\n", s )
#define PII(a,b) printf ( "%d %d", a, b )
#define PIIL(a,b) printf ( "%d %d\n", a, b )
#define PIII(a,b,c) printf ( "%d %d %d", a, b, c )
#define PIIIL(a,b,c) printf ( "%d %d %d\n", a, b, c )
#define rep(i,m,n) for ( int i = m; i <  n; i ++ )
#define REP(i,m,n) for ( int i = m; i <= n; i ++ )
#define dep(i,m,n) for ( int i = m; i >  n; i -- )
#define DEP(i,m,n) for ( int i = m; i >= n; i -- )
#define repi(i,m,n,k) for ( int i = m; i <  n; i += k )
#define REPI(i,m,n,k) for ( int i = m; i <= n; i += k )
#define depi(i,m,n,k) for ( int i = m; i >  n; i += k )
#define DEPI(i,m,n,k) for ( int i = m; i >= n; i -= k )
#pragma comment ( linker, "/STACK:131072000,131072000" )
using namespace std;
template <class T>
inline bool RD ( T &ret )
{
    char c;
    int sgn;
    if ( c = getchar(), c ==EOF )return 0; //EOF
    while ( c != '-' && ( c < '0' || c > '9' ) ) c = getchar();
    sgn = ( c == '-' ) ? -1 : 1;
    ret = ( c == '-' ) ? 0 : ( c - '0' );
    while ( c = getchar() , c >= '0' && c <= '9' ) ret = ret * 10 + ( c - '0' );
    ret *= sgn;
    return 1;
}
inline void PD ( int x )
{
    if ( x > 9 ) PD ( x / 10 );
    putchar ( x % 10 + '0' );
}
struct node{
    unsigned x, y;
};
struct edge{
    long long x, y;
};
const long long mod = 4294967296;
node num[10000005];
bool cmp ( const node &a, const node &b )
{
    return a.x < b.x;
}
int main()
{
    int T;
    scanf ( "%d", &T );
    while ( T -- ){
        int n;
        long long l, r, a, b, c, d;
        RD ( n );RD ( l );RD( r );RD( a );RD( b );RD( c );RD( d );
        num[0].x = l;
        num[0].y = r;
        if ( num[0].x > num[0].y ) swap ( num[0].x, num[0].y );
        for ( int i = 1; i < n; i ++ ){
            l = ( ( l * a ) % mod + b ) % mod;
            r = ( ( r * c ) % mod + d ) % mod;
            num[i].x = l;
            num[i].y = r;
            if ( num[i].x > num[i].y )swap ( num[i].x, num[i].y );
        }
        int flag = 0;
        edge t[4];
        for ( int i = 0; i < 4; i ++ ){
            t[i].x = t[i].y = 9223372036854775807;
        }
        t[0].x = -1;
        t[0].y = -1;
        bool mark = 1;
        for ( int k = 1; k <= 3; k ++ ){
            mark = 0;
            for ( int i = 0; i < n; i ++ ){
                if ( t[k-1].y < num[i].x && t[k].y > num[i].y ){
                    t[k].y = num[i].y;
                    t[k].x = num[i].x;
                    mark = 1;
                }
            }
            if ( mark == 1 )flag ++;
        }
        if ( flag >= 3 ){
            printf ( "YES\n" );
        }
        else{
            printf ( "NO\n" );
        }
    }
    return 0;
}