题解:Dhaka 2011
程序员文章站
2022-03-03 11:36:00
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Binary Matrix
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#define N 2010
using namespace std;
int const INF=(1<<30);
int ask(int *a,int n,int tot)
{
vector<int> v(a,a+n);
int ans=0;
for(int i=0; i+1<v.size();++i)
{
ans+=abs(v[i]-tot);
v[i+1]+=v[i]-tot;
}
return ans;
}
int a[N],b[N],n,m,T,sum;
char s[N];
int main()
{
scanf("%d",&T);
for (int t=1; t<=T; ++t)
{
scanf("%d%d",&n,&m);
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
sum=0;
for (int i=1; i<=n; ++i)
{
scanf("%s",s+1);
for (int j=1; j<=m; ++j)
if (s[j]=='1')
{
++a[i];
++b[j];
++sum;
}
}
int ans1=INF,ans2=INF;
if (sum%n==0)
{
for (int i=1; i<=n; ++i) a[i+n]=a[i];
for (int i=1; i<=n; ++i) ans1=min(ans1,ask(a+i,n,sum/n));
}
if (sum%m==0)
{
for (int i=1; i<=m; ++i) b[i+m]=b[i];
for (int i=1; i<=m; ++i) ans2=min(ans2,ask(b+i,m,sum/m));
}
printf("Case %d: ",t);
if (ans1==INF && ans2==INF) puts("impossible");
else if (ans1==INF) printf("column %d\n",ans2);
else if (ans2==INF) printf("row %d\n",ans1);
else printf("both %d\n",ans1+ans2);
}
}
Candles
#include<bits/stdc++.h>
#define COUNT __builtin_popcount
using namespace std;
bool ask(int l,int r,int i)
{
if(l==0)return i>>r&1;
return l!=r&&(i>>r&1)&&(i>>l&1);
}
int main()
{
for(int kase=0,n,a[15],ans; ~scanf("%d",&n)&&n; printf("\n"))
{
for(int i=0; i<n; ++i)scanf("%d",&a[i]);
for(int i=0,len=ans=(1<<10)-1,flag; i<len; ++i)
if(COUNT(i)<COUNT(ans))
{
for(int j=flag=0,l,r; !flag&&j<n; ++j)
{
l=a[j]/10,r=a[j]%10;
if(ask(l,r,i))continue;
flag=1;
if(l==0)
{
for(int k=1; k+k<r; ++k)
if((i>>k&1)&&(i>>r-k&1))
{
flag=0;
break;
}
}
else
{
for(int k=1,ii; flag&&k<10; ++k)
if(i>>k&1)
{
l=(a[j]-k)/10,r=(a[j]-k)%10,ii=i&~(1<<k);
if(ask(l,r,ii))
{
flag=0;
break;
}
for(int kk=0,iii; kk<10; ++kk)
if(ii>>kk&1)
{
if(a[j]-k*10-kk<0)break;
l=(a[j]-k*10-kk)/10,r=(a[j]-k*10-kk)%10,iii=ii&~(1<<kk);
if(ask(l,r,iii))
{
flag=0;
break;
}
}
}
}
}
if(!flag)ans=i;
}
printf("Case %d: ",++kase);
for(int i=9; ~i; --i)
if(ans>>i&1)printf("%d",i);
}
}
Cards
#include<bits/stdc++.h>
using namespace std;
typedef double lf;
const lf EPS=1e-4,INF=1e4;
int t,a[4],kase;
lf f[15][15][15][15][5][5];
lf dp(int i,int j,int k,int l,int m,int n)
{
if(f[i][j][k][l][m][n]>=-EPS)return f[i][j][k][l][m][n];
lf &ans=f[i][j][k][l][m][n]=0;
int need[4]= {a[0]-i,a[1]-j,a[2]-k,a[3]-l},res=54-i-j-k-l,ca[4]= {13-i,13-j,13-k,13-l};
if(m<4)--need[m],--res;
if(n<4)--need[n],--res;
if(*max_element(need,need+4)<=0)return ans=54-res;
if(ca[0])ans+=dp(i+1,j,k,l,m,n)/res*ca[0];
if(ca[1])ans+=dp(i,j+1,k,l,m,n)/res*ca[1];
if(ca[2])ans+=dp(i,j,k+1,l,m,n)/res*ca[2];
if(ca[3])ans+=dp(i,j,k,l+1,m,n)/res*ca[3];
if(m==4)
{
lf tmp=INF;
for(int t=0; t<4; ++t)
if(need[t]>0)
tmp=min(tmp,dp(i,j,k,l,t,n));
ans+=tmp/res*2;
}
else if(n==4)
{
lf tmp=INF;
for(int t=0; t<4; ++t)
if(need[t]>0)
tmp=min(tmp,dp(i,j,k,l,m,t));
ans+=tmp/res;
}
if(ans<EPS)ans=INF;
return ans;
}
int main()
{
for(scanf("%d",&t); t--;)
{
printf("Case %d: ",++kase);
for(int i=0; i<4; ++i)scanf("%d",&a[i]);
for(int i=0; i<15; ++i)
for(int j=0; j<15; ++j)
for(int k=0; k<15; ++k)
for(int l=0; l<15; ++l)
for(int m=0; m<5; ++m)
for(int n=0; n<5; ++n)
f[i][j][k][l][m][n]=-1;
if(a[0]+a[1]+a[2]+a[3]==0)
{
printf("0.000\n");
continue;
}
if(max(a[0]-13,0)+max(a[1]-13,0)+max(a[2]-13,0)+max(a[3]-13,0)>2)
{
printf("-1.000\n");
continue;
}
printf("%.3f\n",dp(0,0,0,0,4,4));
}
}
Packing for Holiday
#include<bits/stdc++.h>
using namespace std;
int t,a,b,c,kase;
int main()
{
for(scanf("%d",&t); t--;)
{
scanf("%d%d%d",&a,&b,&c);
printf("Case %d: %s\n",++kase,a<21&&b<21&&c<21?"good":"bad");
}
}
As Long as I Learn, I Live
#include <cstdio>
#include <cstring>
using namespace std;
int T,n,m;
int line[200][200];
int go[200],a[200];
int main()
{
scanf("%d",&T);
for (int t=1;t<=T;++t)
{
int ans=0;
memset(line,0,sizeof(line));
memset(go,0,sizeof(go));
scanf("%d%d",&n,&m);
for (int i=0;i<n;++i) scanf("%d",&a[i]);
for (int i=0;i<m;++i)
{
int u,v;
scanf("%d%d",&u,&v);
line[u][v]=true;
++go[u];
}
int now=0;
while (go[now])
{
int maxid=-1,maxa=0;
for (int i=0;i<n;++i) if (line[now][i] && maxa<a[i])
{
maxa=a[i];maxid=i;
}
ans+=maxa;
now=maxid;
}
printf("Case %d: %d %d\n",t,ans,now);
}
}
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