Fork me on GitHub

「HDU5823」2016多校联合Round8 Color II 状压

发表于 | 分类于
字数统计: 516字 | 阅读时长 ≈ 3分钟

Links there:HDU5823

题意

求一个图的每一个子图的最小染色数.

点集大小 $N \leq 18$

思路

暴力枚举子集的复杂度是 $O(3^n)$ ,等等! 似乎 $3^{18} = 387420489$ 可能常数优秀可以过？

我们先枚举子图 再枚举子集 是独立集合 涂一个颜色

然后将这个子集剩下的状态转移出来 $f[S] = min(f[S],f[S补]+1)$

交了一发 居然过了!

其实利用快速沃尔什变化可以优化到 $O(2^N \times N^2)$

具体可以看这里Link

Code

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
//my vegetable has exploded. :(
#include<bits/stdc++.h>
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MM(x,y) memset(x,y,sizeof(x))
#define MCPY(a,b) memcpy(a,b,sizeof(b))
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=b;i>=a;i--)
#define fi first
#define se second
using namespace std;
#define int unsigned
inline int quickpow(int m,int n,int p){int b=1;while(n){if(n&1)b=b*m%p;n=n>>1;m=m*m%p;}return b;}
inline int getinv(int x,int p){return quickpow(x,p-2,p);}
inline int read(void){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){f=ch=='-'?-1:1;ch=getchar();}
    while(isdigit(ch)){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return x * f;
}
int msk,t,n;
char s[19][19];
int res[1<<18],valid[1<<18];
///------------------head------------------
signed main(signed argc, char *argv[]){
    t=read();
    while(t--){
        n=read();
        for (int i = 1; i <= n; ++i) scanf("%s",s[i]+1);
        MM(valid,1); MM(res,0x3F);
        msk = (1 << n) - 1;
        for (int i = 1; i <= msk; ++i){
            for (int j = 1; j <= n; ++j)
            {
                if (i>>(j-1)&1)
                 for (int k = 1; k <= n; ++k)
                if ((i>>(k-1)&1) && s[j][k] == '1') {valid[i] = 0; break;}
                if (!valid[i]) break;
            }
        }
        res[0] = 0;
        for (int i = 1; i <= msk; ++i)
            for (int S = i; S; S = (S-1)&i)
            if (valid[S]) res[i] = min(res[i],res[S^i] + 1);
        int mul = 1,ans = 0;
        for (int i = 1; i <= msk; ++i) mul = mul * 233, ans += res[i] * mul;
        printf("%u\n",ans);
    }
    return 0;
}

/* Examples: */
/*

*/

/*

*/