Redraw Beautiful Drawings
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1660 Accepted Submission(s): 357
Problem Description
Alice and Bob are playing together. Alice is crazy about art and she has visited many museums around the world. She has a good memory and she can remember all drawings she has seen.
Today Alice designs a game using these drawings in her memory. First, she matches K+1 colors appears in the picture to K+1 different integers(from 0 to K). After that, she slices the drawing into grids and there are N rows and M columns. Each grid has an integer
on it(from 0 to K) representing the color on the corresponding position in the original drawing. Alice wants to share the wonderful drawings with Bob and she tells Bob the size of the drawing, the number of different colors, and the sum of integers on each
row and each column. Bob has to redraw the drawing with Alice‘s information. Unfortunately, somtimes, the information Alice offers is wrong because of Alice‘s poor math. And sometimes, Bob can work out multiple different drawings using the information Alice
provides. Bob gets confused and he needs your help. You have to tell Bob if Alice‘s information is right and if her information is right you should also tell Bob whether he can get a unique drawing.
Input
The input contains mutiple testcases.
For each testcase, the first line contains three integers N(1 ≤ N ≤ 400) , M(1 ≤ M ≤ 400) and K(1 ≤ K ≤ 40).
N integers are given in the second line representing the sum of N rows.
M integers are given in the third line representing the sum of M columns.
The input is terminated by EOF.
Output
For each testcase, if there is no solution for Bob, output "Impossible" in one line(without the quotation mark); if there is only one solution for Bob, output "Unique" in one line(without the quotation mark) and output an N * M matrix in the following N lines
representing Bob‘s unique solution; if there are many ways for Bob to redraw the drawing, output "Not Unique" in one line(without the quotation mark).
Sample Input
2 2 4 4 2 4 2 4 2 2 2 2 5 0 5 4 1 4 3 9 1 2 3 3
Sample Output
Not Unique Impossible Unique 1 2 3 3
Author
Fudan University
给你一个n*m的矩阵，然后个格子里面有一个小于k的数，而且告诉你每行和每列的和，让你求是否存在解，并推断是否唯一。
建图：源点和每行连边，容量为每行的和。每列和汇点连边，容量为每列的和。每行和每列连边，容量为k。
推断是否存在唯一解，在残留网络里面是否能找到一个环。
//515MS 7304K
#include<stdio.h>
#include<string.h>
#define M 1007
int s,t,n,m,k,sum1,sum2;
int row[M],col[M],vis[M],g[M][M];
const int MAXN=20010;//点数的最大值
const int MAXM=880010;//边数的最大值
const int INF=0x3f3f3f3f;
struct Node
{
int from,to,next;
int cap;
}edge[MAXM];
int tol;
int head[MAXN];
int dis[MAXN];
int gap[MAXN];//gap[x]=y :说明残留网络中dis[i]==x的个数为y
void init()
{
tol=0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w)
{
edge[tol].from=u;
edge[tol].to=v;
edge[tol].cap=w;
edge[tol].next=head[u];
head[u]=tol++;
edge[tol].from=v;
edge[tol].to=u;
edge[tol].cap=0;
edge[tol].next=head[v];
head[v]=tol++;
}
void BFS(int start,int end)
{
memset(dis,-1,sizeof(dis));
memset(gap,0,sizeof(gap));
gap[0]=1;
int que[MAXN];
int front,rear;
front=rear=0;
dis[end]=0;
que[rear++]=end;
while(front!=rear)
{
int u=que[front++];
if(front==MAXN)front=0;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(dis[v]!=-1)continue;
que[rear++]=v;
if(rear==MAXN)rear=0;
dis[v]=dis[u]+1;
++gap[dis[v]];
}
}
}
int SAP(int start,int end)
{
int res=0,nn=end+1;
BFS(start,end);
int cur[MAXN];
int S[MAXN];
int top=0;
memcpy(cur,head,sizeof(head));
int u=start;
int i;
while(dis[start]<nn)
{
if(u==end)
{
int temp=INF;
int inser;
for(i=0;i<top;i++)
if(temp>edge[S[i]].cap)
{
temp=edge[S[i]].cap;
inser=i;
}
for(i=0;i<top;i++)
{
edge[S[i]].cap-=temp;
edge[S[i]^1].cap+=temp;
}
res+=temp;
top=inser;
u=edge[S[top]].from;
}
if(u!=end&&gap[dis[u]-1]==0)//出现断层，无增广路
break;
for(i=cur[u];i!=-1;i=edge[i].next)
if(edge[i].cap!=0&&dis[u]==dis[edge[i].to]+1)
break;
if(i!=-1)
{
cur[u]=i;
S[top++]=i;
u=edge[i].to;
}
else
{
int min=nn;
for(i=head[u];i!=-1;i=edge[i].next)
{
if(edge[i].cap==0)continue;
if(min>dis[edge[i].to])
{
min=dis[edge[i].to];
cur[u]=i;
}
}
--gap[dis[u]];
dis[u]=min+1;
++gap[dis[u]];
if(u!=start)u=edge[S[--top]].from;
}
}
return res;
}
void build()
{
sum1=0,sum2=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&row[i]);
sum1+=row[i];
addedge(s,i,row[i]);
}
for(int i=1;i<=m;i++)
{
scanf("%d",&col[i]);
sum2+=col[i];
addedge(n+i,t,col[i]);
}
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
addedge(i,n+j,k);
}
bool iscycle(int u,int fa)
{
for(int i=head[u];i!=-1;i=edge[i].next)
{
if(i==(fa^1))continue;
if(edge[i].cap)
{
if(vis[edge[i].to])return true;
vis[edge[i].to]=true;
if(iscycle(edge[i].to,i))return true;
vis[edge[i].to]=false;
}
}
return false;
}
void solve()
{
if(sum1!=sum2){printf("Impossible\n");return;}
int anss=SAP(s,t);
//printf("anss=%d\n",anss);
if(anss!=sum1){printf("Impossible\n");return;}
memset(vis,false,sizeof(vis));
int flag=0;
for(int i=1;i<=n;i++)//推断残留网络是否存在环
if(iscycle(i,-1))
{flag=1;break;}
if(flag)printf("Not Unique\n");
else
{
printf("Unique\n");
for(int u=1;u<=n;u++)
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(v>n&&v<=n+m)
g[u][v-n]=k-edge[i].cap;
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<m;j++)
printf("%d ",g[i][j]);
printf("%d\n",g[i][m]);
}
}
}
int main()
{
while(scanf("%d%d%d",&n,&m,&k)!=EOF)
{
s=0,t=n+m+1;
init();
build();
solve();
}
}