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HDU 4888

2024-07-15 18:40:08   218人阅读

Redraw Beautiful Drawings

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1993    Accepted Submission(s): 446


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
 

Source


已知一个矩阵，n行m列。已知每一行，每一列的和，判断是否存在这样的矩阵，并且判断是否唯一。

很裸的最大流模型，前面的部分很水，关键是，最大流判环部分有点纠结。

代码：

/* ***********************************************
Author :rabbit
Created Time :2014/8/4 18:18:37
File Name :11.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
const int maxn=20100;
const int maxm=1002000;
struct Edge{
	int next,to,cap;
	Edge(){};
	Edge(int _next,int _to,int _cap){
		next=_next;to=_to;cap=_cap;
	}
}edge[maxm];
int head[maxn],tol,dep[maxn],gap[maxn];
void addedge(int u,int v,int flow){
    edge[tol]=Edge(head[u],v,flow);head[u]=tol++;
    edge[tol]=Edge(head[v],u,0);head[v]=tol++;
}
void bfs(int start,int end){
    memset(dep,-1,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0]++;int front=0,rear=0,Q[maxn];
    dep[end]=0;Q[rear++]=end;
    while(front!=rear){
        int u=Q[front++];
        for(int i=head[u];i!=-1;i=edge[i].next){
            int v=edge[i].to;if(dep[v]==-1)
                Q[rear++]=v,dep[v]=dep[u]+1,gap[dep[v]]++;
        }
    }
}
int cur[maxn],S[maxn];
int sap(int s,int t,int N){
	int res=0;bfs(s,t);
	int top=0,u=s,i;
	memcpy(cur,head,sizeof(head));
	while(dep[s]<N){
		if(u==t){
			int temp=INF,id;
		    for( i=0;i<top;i++)
			   if(temp>edge[S[i]].cap)
				   temp=edge[S[i]].cap,id=i;
		    for( i=0;i<top;i++)
			      edge[S[i]].cap-=temp,edge[S[i]^1].cap+=temp;
		    res+=temp;top=id;u=edge[S[top]^1].to;
		}
		if(u!=t&&gap[dep[u]-1]==0)break;
		for( i=cur[u];i!=-1;i=edge[i].next)
			if(edge[i].cap&&dep[u]==dep[edge[i].to]+1)break;
		if(i!=-1)cur[u]=i,S[top++]=i,u=edge[i].to;
		else{
			int MIN=N;
			for( i=head[u];i!=-1;i=edge[i].next)
				if(edge[i].cap&&MIN>dep[edge[i].to])
					MIN=dep[edge[i].to],cur[u]=i;
			--gap[dep[u]];++gap[dep[u]=MIN+1];
			if(u!=s)u=edge[S[--top]^1].to;
		}
	}
	return res;
}
int hang[500],lie[500];
int flow[500][500],vis[maxn],m,n,k;
bool dfs(int u,int fa){
	if(vis[u])return 1;
	vis[u]=1;
	for(int i=head[u];i!=-1;i=edge[i].next){
		int v=edge[i].to;
		if(v==fa||v==0||v==n+m+1||edge[i].cap==0)continue;
		if(dfs(v,u))return 1;
	}
	vis[u]=0;
	return 0;
}
int main()
{
     freopen("1002.in","r",stdin);
     freopen("data.out","w",stdout);
	 while(~scanf("%d%d%d",&n,&m,&k)){
		 memset(head,-1,sizeof(head));tol=0;
		 int L=0,R=0;
		 for(int i=1;i<=n;i++)scanf("%d",&hang[i]),L+=hang[i];
		 for(int i=1;i<=m;i++)scanf("%d",&lie[i]),R+=lie[i];
		 if(L!=R){
			 puts("Impossible");continue;
		 }
		 for(int i=1;i<=n;i++)
			 addedge(0,i,hang[i]);
		 for(int i=1;i<=m;i++)
			 addedge(i+n,n+m+1,lie[i]);
		 for(int i=1;i<=n;i++)
			 for(int j=n+1;j<=n+m;j++)
				 addedge(i,j,k);
		 int ans=sap(0,n+m+1,m+n+100);
		 if(ans!=L){
			 puts("Impossible");continue;
		 }
		 int flag=0;
		 memset(vis,0,sizeof(vis));
		 for(int i=1;i<=n;i++)
			 if(dfs(i,-1)){
				 flag=1;break;
			 }
		 if(flag)puts("Not Unique");
		 else {
			 puts("Unique");
			 for(int i=1;i<=n;i++)
				 for(int j=head[i];j!=-1;j=edge[j].next){
					 int v=edge[j].to;
					 if(v>n&&v<=n+m)
						 flow[i][v-n]=k-edge[j].cap;
				 }
			 for(int i=1;i<=n;i++){
				 printf("%d",flow[i][1]);
				 for(int j=2;j<=m;j++)
					 printf(" %d",flow[i][j]);
				 puts("");
			 }
		 }
	 }
     return 0;
}



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