day4

alien
飞船降落问题
矩阵前缀和+枚举
求出前缀和，暴力枚举每一个子矩阵的和，是0就可以降落
注意有正方形飞船，长度不等于点

#include<cstdio>#include<iostream>#include<cstring>using namespace std;#define D "%d"#define for(i,x,y) for(int i=x;i<=y;i++)#define prf printf#define scf scanfint n,m,r,c,ans;int a[1999][1999],s[1999][1999];int main(){    freopen("alien.in","r",stdin);    freopen("alien.out","w",stdout);    scf(D D D D,&n,&m,&r,&c);    for(i,1,r)    {        int x,y;        scf(D D,&x,&y);        a[x][y]=1;    }    for(i,1,n)     for(j,1,m)         s[i][j]=s[i-1][j]+s[i][j-1]-s[i-1][j-1]+a[i][j];    for(k,1,c)    {        int x,y;        ans=0;        scf(D D,&x,&y);        for(i,1,n)         for(j,1,m)        if(i+x-1<=n&&j+y-1<=m)        {            int w=s[i+x-1][j+y-1]-s[i-1][j+y-1]-s[i+x-1][j-1]+s[i-1][j-1];            if(w==0)            ans++;        }        if(x!=y)        {            for(i,1,n)             for(j,1,m)            if(i+y-1<=n&&j+x-1<=m)            {            int w=s[i+y-1][j+x-1]-s[i-1][j+x-1]-s[i+y-1][j-1]+s[i-1][j-1];            if(w==0)            ans++;            }           }        prf(D "\n",ans);    }}

game
博弈题
有n个点，m条边
两个人给点染色
得到点的权值和边的权值，求最优解的差（保证点偶数个）
直接均摊
他们都是最优解
所以不会看到另一个人得到最多的
所以一定会然一条边的另一边
最后排序减一下就行了

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespace std;#define D "%d"#define for(i,x,y,z) for(int i=x;i<=y;i+=z)#define prf printf#define scf scanfint n,m;double w[19199],ans;int main(){    freopen(".in","r",stdin);    freopen(".out","w",stdout);    scf(D D,&n,&m);    for(i,1,n,1)        scf("%lf",&w[i]);    for(i,1,m,1)    {        int x,y;        double z;        scf(D D "%lf",&x,&y,&z);        w[x]+=z/2.0;        w[y]+=z/2.0;    }    sort(w+1,w+n+1);    for(i,1,n,2)        ans+=w[i+1]-w[i];    prf(D,int (ans));}

fortress
大爆搜

没有什么技巧
就是一个超级大爆搜花了我两小时才调完
先打个表，将读入的问题搞好
再用邻接表弄，没墙就连上边
dfs搜有多少房间，搜最大的就先那种染色题一样
然后再删边添边，一次一次爆搜，找最大的

#include<cstdio>#include<cstring>#include<iostream>using namespace std;int a[99][99],c[2600][2699],f[9999],ans,m,n,x,y;char z;int nex[99999],head[9999],to[9999],tot,max1;int up(int x,int y){    to[tot]=0;    c[y][x]=0;    head[y]=nex[tot];    nex[tot]=0;    tot--;    to[tot]=0;    c[x][y]=0;    head[x]=nex[tot];    nex[tot]=0;    tot--;}int add(int x,int y){    tot++;    to[tot]=y;    c[x][y]=1;    nex[tot]=head[x];    head[x]=tot;    tot++;    to[tot]=x;    c[y][x]=1;    nex[tot]=head[y];    head[y]=tot;}int dfs(int x){    for(int i=head[x];i;i=nex[i])    {        int tmp=to[i];        if(!f[tmp])        {            ans++;            f[tmp]=1;            dfs(tmp);        }    }}int cnt;int b[16][5]={    4,1,2,4,8,    3,2,4,8,0,    3,1,4,8,0,    2,4,8,0,0,    3,1,2,8,0,    2,2,8,0,0,    2,1,8,0,0,    1,8,0,0,0,    3,1,2,4,0,    2,2,4,0,0,    2,1,4,0,0,    1,4,0,0,0,    2,1,2,0,0,    1,2,0,0,0,    1,1,0,0,0,    0,0,0,0,0,};int main(){    freopen("fortress.in","r",stdin);    freopen("fortress.out","w",stdout);    scanf("%d%d",&m,&n);int t=0;    for(int i=1;i<=n;i++)    for(int j=1;j<=m;j++)    {        ++t;        a[i][j]=t;    }    for(int i=1;i<=n;i++)    for(int j=1;j<=m;j++)    {        int x;        scanf("%d",&x);        for(int k=1;k<=b[x][0];k++)        {            if(b[x][k]==2&&i-1>=1)            add(a[i][j],a[i-1][j]);            if(b[x][k]==1&&j-1>=1)            add(a[i][j],a[i][j-1]);            if(b[x][k]==8&&i+1<=n)            add(a[i][j],a[i+1][j]);            if(b[x][k]==4&&j+1<=m)            add(a[i][j],a[i][j+1]);        }    }    for(int i=1;i<=t;i++)    if(!f[i])    f[i]=1,cnt++,ans=1,dfs(i),max1=max(ans,max1);    printf("%d\n%d\n",cnt,max1);max1=0;    for(int j=1;j<=m;j++)    for(int i=n;i>=1;i--)    {        memset(f,0,sizeof(f));        if(!c[a[i][j]][a[i-1][j]]&&i-1>=1)        {            add(a[i][j],a[i-1][j]);            for(int k=1;k<=n;k++)            for(int q=1;q<=m;q++)            {                ans=0;                dfs(a[i][j]);                if(max1<ans)                {                    max1=ans;                    x=i,y=j,z='N';                }            }            up(a[i][j],a[i-1][j]);        }        memset(f,0,sizeof(f));        if(!c[a[i][j]][a[i][j+1]]&&j+1<=m)        {            add(a[i][j],a[i][j+1]);            for(int k=1;k<=n;k++)            for(int q=1;q<=m;q++)            {                ans=0;                dfs(a[i][j]);                if(max1<ans)                {                    max1=ans;                    x=i,y=j,z='E';                }            }            up(a[i][j],a[i][j+1]);        }    }    printf("%d\n%d %d %c",max1,x,y,z);}