SEERC 2017 E.Looping Playlist

题面

E.Looping Playlist

题意

播放列表从任意位置开始循环播放一次, 给定音高, 求最少歌曲数.

每首歌曲有两个(含)以上音符组成, 均遵循某种大调音阶(major scale).

下面给出题目中的12种大调音阶:

major scale	Root +0	Root +2	Root +4	Root +5	Root +7	Root +9	Root +11	Root +12
Do	Do	Re	Mi	Fa	Sol	La	Si	Do
Do#	Do#	Re#	Fa	Fa#	Sol#	La#	Do	Do#
Re	Re	Mi	Fa#	Sol	La	Si	Do#	Re
Re#	Re#	Fa	Sol	Sol#	La#	Do	Re	Re#
Mi	Mi	Fa#	Sol#	La	Si	Do#	Re#	Mi
Fa	Fa	Sol	La	La#	Do	Re	Mi	Fa
Fa#	Fa#	Sol#	La#	Si	Do#	Re#	Fa	Fa#
Sol	Sol	La	Si	Do	Re	Mi	Fa#	Sol
Sol#	Sol#	La#	Do	Do#	Re#	Fa	Sol	Sol#
La	La	Si	Do#	Re	Mi	Fa#	Sol#	La
La#	La#	Do	Re	Re#	Fa	Sol	La	La#
Si	Si	Do#	Re#	Mi	Fa#	Sol#	La#	Si

基本思路

首先研究一组大调音阶, 显然, 对给定的一组大调音阶, 可以存在的音符是确定的, 而在音高序列中某种音符只有两种状态(存在和不存在), 很自然我们想到用二进制来处理.

接下来很重要的一步就是将音高序列转换为二进制数, 赋予每种音符一定的权值, 然后移位 + 按位或即可, 不过多赘述.

那么, 如何确定音高序列和大调音阶是否匹配呢? 很显然, 如果匹配, 二者按位或的结果中为1的bit数必然等于7(想一想, 为什么?)

于是我们就可以使用贪心来处理了, 遍历每个音符, 然后判断是否有匹配的大调音阶, 一旦加入某个音符后无法匹配, 说明这个音符属于另一个大调音阶, 也就是另一首歌曲.

因为是循环播放, 所以我们枚举开始音符(长度为12的区间就可以了, 自行证明)来计算歌曲数的最小值, 最后还需要考虑结尾和开始(长度可以压缩到14, 或者更小)是否是同一首歌.

打表

有了上面的性质, 我们就可以打表了, 逐一判断是否有匹配的序列就可以了, 然而这样朴素的打表会TLE, 于是我们还需要优化.

我们至多用到了 12 bit, 所以最多有 4095 种情况, 打表是可以接受的, 于是只要遍历然后判断是否符合某个大调音阶就可以了.

打表的相关代码如下(如有需要请自行修改):

点我查看完整代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<bitset<32>> vbi;
typedef bitset<32> bi;
typedef map<string, int> msi;
msi a = {{"Do", 0}, {"Do#", 1}, {"Re", 2}, {"Re#", 3}, {"Mi", 4}, {"Fa", 5}, {"Fa#", 6}, {"Sol", 7}, {"Sol#", 8}, {"La", 9}, {"La#", 10}, {"Si", 11}};
string m[] = {"Do", "Do#", "Re", "Re#", "Mi", "Fa", "Fa#", "Sol", "Sol#", "La", "La#", "Si"};
int ord[] = {0, 2, 4, 5, 7, 9, 11, 12};
vbi f = {2741, 1387, 2774, 1453, 2906, 1717, 3434, 2773, 1451, 2902, 1709, 3418};
bool s[5001] = {0};
const int count1 = 7;
void sc(int n){
    int now = n;
    int ans = 0;
    for (int i = 0; i < 8; i++){
        now = n + ord[i];
        if (now > 11){
            now %= 12;
        }
        ans |= (1 << now);
        //cout << m[now] << ","; 
    }
    bitset<32> bin(ans);
    printf("%5d : ", ans);
    cout << bin.to_string() << '\n';
}
int pd(int t){
    for (int i = 0; i < 12; i++){
        int n = (int)f[i].to_ulong() | t;
        bi temp(n);
        if (temp.count() == count1){
            //printf("%d\n", i);
            return 1;
        }
    }
    return 0;
}
int main(){
    /*for (int i = 0; i < 12; i++){
        sc(i);
    }*/
    //freopen("SEERC_2017-E.in", "w", stdout);
    for (int i = 1; i <= 5000; i++){
        if (pd(i)){
            s[i] = 1;
        }
    }
    for (int i = 0; i <= 5000; i++){
        printf("%d,", s[i]);
    }
    return 0;
} 

完整AC代码

事实证明, 优化过的cin读入比scanf()读入要快, 前者勉强AC(1647ms ~ 2s), 后者TLE.

点我查看完整代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<bitset<32>> vbi;
typedef bitset<32> bi;
typedef map<string, int> msi;
msi yf = {{"Do", 0}, {"Do#", 1}, {"Re", 2}, {"Re#", 3}, {"Mi", 4}, {"Fa", 5}, {"Fa#", 6}, {"Sol", 7}, {"Sol#", 8}, {"La", 9}, {"La#", 10}, {"Si", 11}};
vbi f = {2741, 1387, 2774, 1453, 2906, 1717, 3434, 2773, 1451, 2902, 1709, 3418};
bool pd[] = 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,0};
const int count1 = 7;
const int maxn = 2e7 + 9;
int a[maxn];
/*int pd(int t){
    for (int i = 0; i < 12; i++){
        int n = (int)f[i].to_ulong() | t;
        bi temp(n);
        if (temp.count() == count1){
            //printf("%d\n", i);
            return 1;
        }
    }
    return 0;
}*/
int judge(int l, int r){
    int temp = 0;
    for (int i = l; i <= r; i++){
        temp |= (1 << a[i]);
        if (!pd[temp]){
            return 0;
        }
    }
    return 1;
}
char yinfu[10];
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    //scanf("%d ", &n);
    string s;
    int ans = 0xfffffff;
    int temp = 0;
    for (int i = 1; i <= n; i++){
        //scanf("%s", yinfu);
        //s = yinfu;
        cin >> s;
        a[i] = yf[s];
        a[i + n] = yf[s];
    }
    for (int i = 1; i <= min(12, n); i++){
        int cur = 1, r = 0, l = 0;
        for (int j = i; j <= i + n; j++){
            temp |= (1 << a[j]);
            if (!pd[temp]){
                temp = (1 << a[j]);
                if (!l){
                    l = j - 1;
                }
                cur++;
                r = j;
            }
        }
        if (r){
            r = i + n - r;
            r = min(r, 7);
            if (judge(i + n - r, n + min(i + 7, l))){
                cur--;
            }
        }
        ans = min(cur, ans);
    }
    cout << ans;
    //printf("%d", ans);
    return 0;
}