NOIP 模拟 $24; m matrix$

题解 (by;zjvarphi)

发现 ( m n,m) 都很小，考虑分行状压。

但是上一行和下一行的按钮状态会对当前行造成影响，所以再枚举一个上一行的按钮状态。

因为对于两行，只有如下三种情况是合法的

[0;1;1\ 1;1;0 ]

所以总复杂度为 (mathcal O(n2^n3^n))，最后统计答案时记得最后一行没有下一行来覆盖它，所以它自身的覆盖情况一定要覆盖全。

Code:

#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i
using namespace std;
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++;
    template<typename T>inline void read(T &x) {
        ri f=1;x=0;register char ch=getchar();
        while(ch<'0'||ch>'9') {if (ch=='-') f=0;ch=getchar();};
        while(ch>='0'&&ch<='9') {x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
        x=f?x:-x;
    }
}
using IO::read;
namespace nanfeng{
    #define FI FILE *IN
    #define FO FILE *OUT
    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
    static const int N=11,INF=1061109567;
    int dp[N][1<<N][1<<N],g[N][1<<N],vis[N],mo[N],n,m,ans=INT_MAX;
    char s[N+2];
    inline int main() {
        //FI=freopen("nanfeng.in","r",stdin);
        //FO=freopen("nanfeng.out","w",stdout);
        read(n),read(m);
        for (ri i(1);i<=n;p(i)) {
            scanf("%s",s+1);
            for (ri j(1);j<=m;p(j)) vis[i]|=(s[j]=='1')<<j-1;
        }
        int bs=(1<<m)-1;
        for (ri i(1);i<=n;p(i)) {
            for (ri j(1);j<=m;p(j)) read(mo[j]);
            for (ri j(0);j<=bs;p(j)) {
                ri tmp=0;
                for (ri k(0);k<m;p(k)) if ((j>>k)&1) tmp+=mo[k+1];
                g[i][j]=tmp;
            }
        }
        memset(dp,0x3f,sizeof(dp));
        memset(dp[0],0,sizeof(dp[0]));
        for (ri i(0);i<=bs;p(i)) {
            int tmp=vis[1]|((i<<1|i>>1)&bs)|i;
            dp[1][tmp][i]=g[1][i];
        }
        for (ri i(1);i<n;p(i)) {
            for (ri j(0);j<=bs;p(j)) {
                for (ri l(0);l<=bs;p(l)) {
                    if ((l|j)!=bs) continue;
                    for (ri k(0);k<=bs;p(k)) {
                        if (dp[i][j][k]==INF) continue;
                        int tmp=vis[i+1]|k|((l<<1|l>>1)&bs)|l;
                        dp[i+1][tmp][l]=cmin(dp[i+1][tmp][l],dp[i][j][k]+g[i+1][l]);
                    }
                } 
            }
        }
        for (ri i(0);i<=bs;p(i)) ans=cmin(ans,dp[n][bs][i]);
        printf("%d
",ans);
        return 0;    
    }
}
int main() {return nanfeng::main();}