HDOJ 4975 A simple Gaussian elimination problem.
和HDOJ4888是一样的问题,最大流推断多解
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A simple Gaussian elimination problem.
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1170 Accepted Submission(s): 377
the table after that.
However Dragon's mom came back and found what he had done. She would give dragon a feast if Dragon could reconstruct the table, otherwise keep Dragon hungry. Dragon is so young and so simple so that the original numbers in the table are one-digit number (e.g.
0-9).
Could you help Dragon to do that?
There are three lines for each block. The first line contains two integers N(<=500) and M(<=500), showing the number of rows and columns.
The second line contains N integer show the sum of each row.
The third line contains M integer show the sum of each column.
output one line contains only: "So young!", otherwise (only one way to reconstruct the table) you should output: "So simple!".
3
1 1
5
5
2 2
0 10
0 10
2 2
2 2
2 2
Case #1: So simple!
Case #2: So naive!
Case #3: So young!
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm> #pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; const int maxn=20000;
const int maxm=500000;
const int INF=0x3f3f3f3f; struct Edge
{
int to,next,cap,flow;
}edge[maxm]; int Size,Adj[maxn];
int gap[maxn],dep[maxn],pre[maxn],cur[maxn]; void init()
{
Size=0; memset(Adj,-1,sizeof(Adj));
} void addedge(int u,int v,int w,int rw=0)
{
edge[Size].to=v; edge[Size].cap=w; edge[Size].next=Adj[u];
edge[Size].flow=0; Adj[u]=Size++;
edge[Size].to=u; edge[Size].cap=rw; edge[Size].next=Adj[v];
edge[Size].flow=0; Adj[v]=Size++;
} int sap(int start,int end,int N)
{
memset(gap,0,sizeof(gap));
memset(dep,0,sizeof(dep));
memcpy(cur,Adj,sizeof(Adj)); int u=start;
pre[u]=-1; gap[0]=N;
int ans=0; while(dep[start]<N)
{
if(u==end)
{
int Min=INF;
for(int i=pre[u];~i;i=pre[edge[i^1].to])
if(Min>edge[i].cap-edge[i].flow)
Min=edge[i].cap-edge[i].flow;
for(int i=pre[u];~i;i=pre[edge[i^1].to])
{
edge[i].flow+=Min;
edge[i^1].flow-=Min;
}
u=start;
ans+=Min;
continue;
}
bool flag=false;
int v;
for(int i=cur[u];~i;i=edge[i].next)
{
v=edge[i].to;
if(edge[i].cap-edge[i].flow&&dep[v]+1==dep[u])
{
flag=true;
cur[u]=pre[v]=i;
break;
}
}
if(flag)
{
u=v;
continue;
}
int Min=N;
for(int i=Adj[u];~i;i=edge[i].next)
if(edge[i].cap-edge[i].flow&&dep[edge[i].to]<Min)
{
Min=dep[edge[i].to];
cur[u]=i;
}
gap[dep[u]]--;
if(!gap[dep[u]]) return ans;
dep[u]=Min+1;
gap[dep[u]]++;
if(u!=start) u=edge[pre[u]^1].to;
}
return ans;
}
int n,m;
int a[maxn],b[maxn]; bool vis[maxn],no[maxn];
int Stack[maxm],stk; bool dfs(int u,int pre,bool flag)
{
vis[u]=true;
Stack[stk++]=u;
for(int i=Adj[u];~i;i=edge[i].next)
{
int v=edge[i].to;
if(v==pre) continue;
if(edge[i].flow>=edge[i].cap) continue;
if(!vis[v])
{
if(dfs(v,u,edge[i^1].cap>edge[i^1].flow)) return true;
}
else if(!no[v]) return true;
}
if(flag==false)
{
while(true)
{
int v=Stack[--stk];
no[v]=true;
if(v==u) break;
}
}
return false;
} int main()
{
int T_T,cas=1;
scanf("%d",&T_T);
while(T_T--)
{
scanf("%d%d",&n,&m);
int sum1=0,sum2=0;
for(int i=1;i<=n;i++)
{
scanf("%d",a+i); sum1+=a[i];
}
for(int i=1;i<=m;i++)
{
scanf("%d",b+i); sum2+=b[i];
}
if(sum1!=sum2)
{
printf("Case #%d: So naive!\n",cas++);
continue;
}
if(sum1==sum2&&((sum1==0)||(sum1==n*m*9)))
{
printf("Case #%d: So simple!\n",cas++);
continue;
} /*************build graph*****************/
init();
for(int i=1;i<=n;i++) addedge(0,i,a[i]);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
addedge(i,n+j,9);
for(int i=1;i<=m;i++) addedge(i+n,n+m+1,b[i]);
/*************build graph*****************/
int MaxFlow=sap(0,n+m+1,n+m+2); if(MaxFlow!=sum1)
{
printf("Case #%d: So naive!\n",cas++);
continue;
}
stk=0;
memset(vis,0,sizeof(vis));
memset(no,0,sizeof(no));
if(dfs(n+m+1,n+m+1,0))
{
printf("Case #%d: So young!\n",cas++);
}
else
{
printf("Case #%d: So simple!\n",cas++);
}
}
return 0;
}
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