2016 计蒜之道 热身赛 题解(待续)

硬币翻转

在一个 n 行 m 列的网格中，每个网格中有一枚硬币，在翻转某一个网格中的硬币时，与该网格共边的四个网格中的硬币也将被翻转。求至少需要进行多少次翻转操作，可以将所有的硬币变为正面朝上。
1≤n≤100;1≤m≤15

#include<bits/stdc++.h>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())typedef long long ll;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} #define MAXN (100+10)#define MAXM (15)int a[MAXN],n,m,f[MAXN];bool inside(int x,int y) {    return 1<=x&&x<=n&&1<=y&&y<=m;}int dir[4][2]={{0,1},{0,-1},{-1,0},{1,0}};void clc(int x,int y) {    Rep(di,4) {        int nowx=dir[di][0]+x,nowy=dir[di][1]+y;        if (inside(nowx,nowy)) f[x]^=(1<<y);    }}int main(){//  freopen("A.in","r",stdin);//  freopen(".out","w",stdout);    while(cin>>n>>m) {        int tot=INF;        For(i,n) {            a[i]=0;            For(j,m) a[i]=(a[i]<<1)+read();        }        a[n+1]=0;        Rep(S,1<<m) {            memcpy(f,a,sizeof(a));            int P=S,ans=0;            P=(P^((P<<1)%(1<<m))^(P>>1));            f[1]^=P; f[2]^=S;            ans+=__builtin_popcount(S);             Fork(i,2,n) {                P=f[i-1];                P^=((1<<m)-1);                f[i+1]^=P; f[i-1]^=P;ans+=__builtin_popcount(P);                            P=(P^((P<<1)%(1<<m))^(P>>1));                f[i]^=P;                        }            if (f[n]==((1<<m) -1 ) ) tot=min(ans,tot);        }        if (tot<INF) cout<<tot<<endl;else puts("no solution");    }    return 0;}