Power Stations
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1721 Accepted Submission(s): 469
Special Judge
are connected by cables directly to this town. However, there are some strange bugs in the electric system –One town can only receive electric power from no more than one power station, otherwise the cables will be burned out for overload.
The power stations cannot work all the time. For each station there is an available time range. For example, the power station located on Town 1 may be available from the third day to the fifth day, while the power station on Town 2 may be available from the
first day to the forth day. You can choose a sub-range of the available range as the working time for each station. Note that you can only choose one sub-range for each available range, that is, once the station stops working, you cannot restart it again.
Of course, it is possible not to use any of them.
Now you are given all the information about the cable connection between the towns, and all the power stations’ available time. You need to find out a schedule that every town will get the electricity supply for next D days, one and only one supplier for one
town at any time.
Each of the next M lines contains two integers a, b (1 <= a, b <= N), which means that Town a and Town b are connected directly. Then N lines followed, each contains two numbers si and ei, (1 <= si <= ei <= D) indicating that the available time of Town i’s
power station is from the si-th day to the ei-th day (inclusive).
= 0.
If the plan doesn’t exist, output one line contains “No solution” instead.
Note that the answer may not be unique. Any correct answers will be OK.
Output a blank line after each case.
3 3 5 1 2 2 3 3 1 1 5 1 5 1 5 4 4 5 1 2 2 3 3 4 4 1 1 5 1 5 1 5 1 5
1 5 0 0 0 0 No solution
Statistic | Submit | Discuss | Note
特别注意。。。这题有重边。因为这个让我RE得蛋都残了。。。
然后就是拍dancing links的模版啦
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <queue>
using namespace std;
const int N=60*20*60*6,INF=0xffffff;
int n,m;
int U[N],D[N],L[N],R[N];
int row[N],col[N];
int S[N],H[N],ans[N];
int siz;
int node[66][66],mp[66][66];
int st[66],et[66],a,d,num,np[6][6];
int ans1[N],ans2[N],nps[N],npe[N];
/*
int mas[4444][4444];
void print()
{
memset(mas,0,sizeof(mas));
for(int i=m+1;i<siz;i++)
{
mas[row[i]][col[i]]=1;
}
for(int i=0;i<n;i++)
{
if(i%num==0) cout<<endl;
cout<<i%num<<": \t";
for(int j=1;j<=m;j++)
{
cout<<mas[i][j];
if(j%d==a) cout<<" ";
}
cout<<endl;
}
cout<<endl;
}
*/
void init()
{
for(int i=0;i<=m;i++)
{
L[i]=i-1;
R[i]=i+1;
col[i]=i;
U[i]=D[i]=i;
row[i]=-1;
}
R[m]=0;
L[0]=m;
siz=m+1;
memset(H,0,sizeof(H));
memset(S,0,sizeof(S));
S[0]=INF+1;
}
void insert(int r,int c)
{
U[siz]=U[c];
D[siz]=c;
U[D[siz]]=siz;
D[U[siz]]=siz;
if(H[r])
{
L[siz]=L[H[r]];
R[siz]=H[r];
L[R[siz]]=siz;
R[L[siz]]=siz;
}
else H[r]=L[siz]=R[siz]=siz;
row[siz]=r;
col[siz]=c;
S[c]++;
siz++;
}
void build()
{
init();
for(int i=0;i<a;i++)
{
int s,e,j;
insert(i*num,i+1);
for(s=st[i];s<=et[i];s++)
{
for(e=s;e<=et[i];e++)
{
for(j=s;j<=e;j++)
{
insert(i*num+np[s][e],a+i*d+j+1);
for(int k=1;k<node[i][0];k++) insert(i*num+np[s][e],a+node[i][k]*d+j+1);
}
insert(i*num+np[s][e],i+1);
}
}
}
}
void remove(int c)
{
L[R[c]]=L[c];
R[L[c]]=R[c];
for(int i=D[c];i!=c;i=D[i])
{
for(int j=R[i];j!=i;j=R[j])
{
U[D[j]]=U[j];
D[U[j]]=D[j];
S[col[j]]--;
}
}
}
void resume(int c)
{
for(int i=U[c];i!=c;i=U[i])
{
for(int j=L[i];j!=i;j=L[j])
{
D[U[j]]=j;
U[D[j]]=j;
S[col[j]]++;
}
}
R[L[c]]=c;
L[R[c]]=c;
}
bool dfs(int k)
{
if(R[0]==0) return true;
int mins=INF,c=0;
for(int t=R[0];t!=0;t=R[t])
{
if(S[t]<mins)
{
mins=S[t];
c=t;
}
}
if(c==0) return false;
remove(c);
for(int i=D[c];i!=c;i=D[i])
{
ans[k]=row[i];
for(int j=R[i];j!=i;j=R[j]) remove(col[j]);
if(dfs(k+1)) return true;
for(int j=L[i];j!=i;j=L[j]) resume(col[j]);
}
resume(c);
return false;
}
int main()
{
int b;
ios::sync_with_stdio(false);
while(scanf("%d %d %d",&a,&b,&d)==3)
{
int i,j,k;
num=d*(d+1)/2+1;
n=a*num;
m=a*d+a;
for(k=1,i=0;i<d;i++) for(j=i;j<d;j++) np[i][j]=k++;
memset(mp,0,sizeof(mp));
for(i=0;i<b;i++)
{
scanf("%d %d",&j,&k);
j--;k--;
if(j<k) swap(j,k);
mp[j][k]=1;
}
for(i=0;i<a;i++) node[i][0]=1;
for(j=0;j<a;j++) for(k=0;k<a;k++) if(mp[j][k])
{
node[j][node[j][0]++]=k;
node[k][node[k][0]++]=j;
}
for(i=0;i<a;i++)
{
scanf("%d %d",&st[i],&et[i]);
st[i]--;et[i]--;
}
build();
//print();
if(!dfs(0))
{
puts("No solution\n");
continue;
}
for(i=0;i<a;i++)
{
j=ans[i]%num;
if(j){for(int s=0;s<d;s++) for(int e=s;e<d;e++) if(np[s][e]==j){ans1[ans[i]/num]=s+1;ans2[ans[i]/num]=e+1;}}
else{ans1[ans[i]/num]=ans2[ans[i]/num]=0;}
}
for(i=0;i<a;i++) printf("%d %d\n",ans1[i],ans2[i]);
puts("");
}
return 0;
}