NOIP 模拟 $28; m 遗忘之祭仪$

题解 (by;zjvarphi)

直接贪心模拟即可，对于每个点，如果它未被覆盖，直接在这覆盖一次。

每个黑点只会被扫一次，所以总复杂度为 (mathcal O m (nm))

Code

%: pragma GCC optimize("O9")
%: pragma GCC optimize("inline")
#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,OPUT[100];
    #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(!isdigit(ch)) {if (ch=='-') f=0;ch=getchar();}
        while(isdigit(ch)) {x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
        x=f?x:-x;
    }
    template<typename T>inline void print(T x,char t) {
        if (x<0) putchar('-'),x=-x;
        if (!x) return putchar('0'),(void)putchar(t);
        ri cnt(0);
        while(x) OPUT[p(cnt)]=x%10,x/=10;
        for (ri i(cnt);i;--i) putchar(OPUT[i]^48);
        return (void)putchar(t);
    }
}
using IO::read;using IO::print;
namespace nanfeng{
    #define node(x,y) (node){x,y}
    #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=1e3+7;
    char s[N];
    int mt[N][N],mxx,mxy,mn,cnt,T,n,m,a,b,fg;
    struct node{int x,y;}pnt[N*N];
    inline int check(int x,int y) {
        for (ri i(1);i<=cnt;p(i)) {
            int cx=x+pnt[i].x,cy=y+pnt[i].y;
            if (cx<1||cx>n||cy<1||cy>m) return 0;
            if (!mt[cx][cy]) return 0;
            mt[cx][cy]=0; 
        }
        return 1;
    }
    inline int main() {
        //FI=freopen("nanfeng.in","r",stdin);
        //FO=freopen("nanfeng.out","w",stdout);
        read(T);
        for (ri z(1);z<=T;p(z)) {
            read(n),read(m),read(a),read(b);
            cnt=fg=0;
            mxx=INT_MAX,mxy=INT_MAX;
            for (ri i(1);i<=n;p(i)) {
                scanf("%s",s+1);
                for (ri j(1);j<=m;p(j)) mt[i][j]=(s[j]=='x');
            }
            for (ri i(1);i<=a;p(i)) {
                scanf("%s",s+1);
                for (ri j(1);j<=b;p(j)) 
                    if (s[j]=='x') {
                        pnt[p(cnt)]=node(i,j);
                        if (i<mxx) mxx=i,mxy=j,mn=cnt;
                        else if (i==mxx) if (j<mxy) mxy=j,mn=cnt;
                    }
            }
            if (!cnt) {puts("No");continue;}
            for (ri i(1);i<=cnt;p(i)) {
                if (mn==i) continue;
                pnt[i].x-=pnt[mn].x,pnt[i].y-=pnt[mn].y;
            }
            pnt[mn].x=pnt[mn].y=0;
            for (ri i(1);i<=n&&!fg;p(i)) 
                for (ri j(1);j<=m&&!fg;p(j)) 
                    if (mt[i][j]) if (!check(i,j)) fg=1,puts("No");
            if (fg) continue;
            puts("Yes");
        }
        return 0;
    }
}
int main() {return nanfeng::main();}