$POJ2279\ Mr.Young's\ Picture\ Permutations\ dp$

正解:$dp$

解题报告:

传送门!

先放下题目大意昂$QwQ$,就说给定$N$个人,$K$排,以及每排的人数$a_{1},...,a_{K}$,现要求每排人数递减,然后每列人数也递减,求排列方案数

显然考虑从矮到高考虑,就有一定要是满足轮廓是个凸的,就长得有点儿像之前寒假考试的搜索专题D2T5(记搜)

这题就差不多套路,考虑五维$dp$:设$f_{d_1,d_2,d_3,d_4,d_5}$,然后判断$d_1\leqslant d_2$这种之类乱七八糟的条件,然后瞎转一下,就欧克了

嗷对辣,就是,因为如果直接开$f_{30,30,30,30,30}$会炸空间,,,所以这里要对着读入开,,,啊呀说不清楚还是看$code$趴$QwQ$

 

#include<algorithm>
#include<iostream>
#include<iomanip>
#include<cstring>
#include<cstdio>
#include<map>
using namespace std;
#define il inline
#define gc getchar()
#define int long long
#define ri register int
#define rb register bool
#define rc register char
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define my(i,x,y) for(ri i=x;i>=y;--i)

const int M=5+10,N=1e7;
int n,as,a[M];
bool gdgs=1;

il int read()
{
    rc ch=gc;ri x=0;rb y=1;
    while(ch!='-' && (ch>'9' || ch<'0'))ch=gc;
    if(ch=='-')ch=gc,y=0;
    while(ch>='0' && ch<='9')x=(x<<1)+(x<<3)+(ch^'0'),ch=gc;
    return y?x:x;
}

signed main()
{
    while(gdgs)
    {
        ri n=read();if(!n)return 0;rp(i,1,n)a[i]=read();rp(i,n+1,5)a[i]=0;
        int f[a[1]+1][a[2]+1][a[3]+1][a[4]+1][a[5]+1];memset(f,0,sizeof(f));f[0][0][0][0][0]=1;
        rp(i,1,a[1])
        {
            rp(j,0,min(a[2],i))
            {
                rp(k,0,min(a[3],j))
                {
                    rp(p,0,min(a[4],k))
                    {
                        rp(q,0,min(a[5],p))
                        {
                            if(q)f[i][j][k][p][q]+=f[i][j][k][p][q-1];
                            if(p-1>=q)f[i][j][k][p][q]+=f[i][j][k][p-1][q];
                            if(k-1>=p)f[i][j][k][p][q]+=f[i][j][k-1][p][q];
                            if(j-1>=k)f[i][j][k][p][q]+=f[i][j-1][k][p][q];
                            if(i-1>=j)f[i][j][k][p][q]+=f[i-1][j][k][p][q];
                        }
                    }
                }
            }
        }
        printf("%lld\n",f[a[1]][a[2]][a[3]][a[4]][a[5]]);
    }
    return 0;
}

View Code