和HDOJ4888是一样的问题,最大流推断多解
1.把ISAP卡的根本出不来结果,仅仅能把全为0或者全为满流的给特判掉......
2.在残量网络中找大于2的圈要用一种类似tarjian的方法从汇点開始找,推断哪些点没有到汇点
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
Problem Description
Dragon is studying math. One day, he drew a table with several rows and columns, randomly wrote numbers on each elements of the table. Then he counted the sum of each row and column. Since he thought the map will be useless after he got the sums, he destroyed 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?
Input
The first line of input contains only one integer, T(<=30), the number of test cases. Following T blocks, each block describes one test case.
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
Each output should occupy one line. Each line should start with "Case #i: ", with i implying the case number. For each case, if we cannot get the original table, just output: "So naive!", else if we can reconstruct the table by more than one ways, you should output one line contains only: "So young!", otherwise (only one way to reconstruct the table) you should output: "So simple!".
Sample Input
3
1 1
5
5
2 2
0 10
0 10
2 2
2 2
2 2
Sample Output
Case #1: So simple!
Case #2: So naive!
Case #3: So young!
Author
BJTU
#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;
}
