bzoj 1079 着色方案

题目大意:

有n个木块排成一行,从左到右依次编号为1~n 有k种颜色的油漆,其中第i种颜色的油漆足够涂ci个木块

所有油漆刚好足够涂满所有木块,即c1+c2+...+ck==n

统计任意两个相邻木块颜色不同的着色方案

思路:

记忆化dfs

记录剩余1个能涂,2个能涂……的情况以及前一个的颜色

然后这种颜色-1不能被取到,因为上次它被取了,就减了一

其余的正常的dfs转移

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cmath>
 5 #include<cstdlib>
 6 #include<cstring>
 7 #include<queue>
 8 #include<map>
 9 #include<vector>
10 #define ll long long
11 #define inf 2147483611
12 #define MAXN 17
13 #define MOD 1000000007
14 using namespace std;
15 inline ll read()
16 {
17     ll x=0,f=1;char ch=getchar();
18     while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
19     while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
20     return x*f;
21 }
22 ll n,dp[MAXN][MAXN][MAXN][MAXN][MAXN][7],t[10];
23 ll dfs(int t1,int t2,int t3,int t4,int t5,int ban)
24 {
25     ll& res=dp[t1][t2][t3][t4][t5][ban];
26     if(dp[t1][t2][t3][t4][t5][ban]) return dp[t1][t2][t3][t4][t5][ban];
27     if(t1+t2+t3+t4+t5==0) return 1;
28     if(t1&&ban==2) (res+=(t1-1)*dfs(t1-1,t2,t3,t4,t5,1))%=MOD;
29     if(t1&&ban!=2) (res+=t1*dfs(t1-1,t2,t3,t4,t5,1))%=MOD;
30     if(t2&&ban==3) (res+=(t2-1)*dfs(t1+1,t2-1,t3,t4,t5,2))%=MOD;
31     if(t2&&ban!=3) (res+=t2*dfs(t1+1,t2-1,t3,t4,t5,2))%=MOD;
32     if(t3&&ban==4) (res+=(t3-1)*dfs(t1,t2+1,t3-1,t4,t5,3))%=MOD;
33     if(t3&&ban!=4) (res+=t3*dfs(t1,t2+1,t3-1,t4,t5,3))%=MOD;
34     if(t4&&ban==5) (res+=(t4-1)*dfs(t1,t2,t3+1,t4-1,t5,4))%=MOD;
35     if(t4&&ban!=5) (res+=t4*dfs(t1,t2,t3+1,t4-1,t5,4))%=MOD;
36     if(t5) (res+=t5*dfs(t1,t2,t3,t4+1,t5-1,5))%=MOD;
37     //cout<<t1<<" "<<t2<<" "<<t3<<" "<<t4<<" "<<t5<<" "<<ban<<" "<<res<<" "<<dp[t1][t2][t3][t4][t5][ban]<<endl;
38     return res;
39 }
40 int main()
41 {
42     n=read();ll a;
43     for(int i=1;i<=n;i++) a=read(),t[a]++;
44     printf("%lld",dfs(t[1],t[2],t[3],t[4],t[5],0));
45 }
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posted @ 2017-11-24 21:12  jack_yyc  阅读(...)  评论(... 编辑 收藏