POJ P2279 Mr. Young's Picture Permutations 题解

每日一题 day14 打卡

Analysis

五维dp
f[a1,a2,a3,a4,a5]表示各排从左端起分别占了a1,a2,a3,a4,a5个人时合影方案数量
然后我们枚举a1,a2,a3,a4,a5从0开始到N1,N2……N5
若a1 < N1
若a2 < N2&a1 > a2
若a3 < N3&a2 > a3
……(以此类推)

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define int long long
 #define maxn 30+10
 #define maxk 5+10
 using namespace std;
 inline int read()
 {
     int x=;
     bool f=;
     char c=getchar();
     for(; !isdigit(c); c=getchar()) if(c=='-') f=;
     for(; isdigit(c); c=getchar()) x=(x<<)+(x<<)+c-'';
     if(f) return x;
     return -x;
 }
 inline void write(int x)
 {
     if(x<){putchar('-');x=-x;}
     if(x>)write(x/);
     putchar(x%+'');
 }
 int k;
 int a[maxk],s[maxk];
 int dp[maxn][maxn][maxn][maxn][maxn];
 signed main()
 {
     while()
     {
         memset(dp,,sizeof(dp));
         memset(a,,sizeof(a));
         k=read();
         if(k==) return ;
         for(int i=;i<=k;i++) a[i]=read();
         dp[][][][][]=;
         for(s[]=;s[]<=a[];s[]++)
             for(s[]=;s[]<=a[];s[]++)
                 for(s[]=;s[]<=a[];s[]++)
                     for(s[]=;s[]<=a[];s[]++)
                         for(s[]=;s[]<=a[];s[]++)
                             for(int i=;i<=k;i++)
                             {
                                 if(s[i]<a[i]&&(i==||s[i-]>s[i]))
                                 {
                                     int x1=s[],x2=s[],x3=s[],x4=s[],x5=s[];
                                     s[i]++;
                                     dp[s[]][s[]][s[]][s[]][s[]]+=dp[x1][x2][x3][x4][x5];
                                     s[i]--;
                                 }
                             }
         write(dp[a[]][a[]][a[]][a[]][a[]]);
         printf("\n");
     }
     return ;
 }

请各位大佬斧正（反正我不认识斧正是什么意思）