HDU 4888 Redraw Beautiful Drawings （最大流）

HDU 4888

Redraw Beautiful Drawings

Redraw Beautiful Drawings

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1856 Accepted Submission(s): 415

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

用最大流去判断有没有解。 判断多解的话，在残余网络中看能不能找到长度大于2的环。  

/* ***************

Author        :kuangbin

Created Time  :2014/8/3 13:06:32

File Name     :E:\2014ACM\比赛\2014多校训练\2014多校3\HDU4888.cpp

************************************************ */

#include <stdio.h>

#include <string.h>

#include <iostream>

#include <algorithm>

#include <vector>

#include <queue>

#include <set>

#include <map>

#include <string>

#include <math.h>

#include <stdlib.h>

#include <time.h>

using namespace std;

const int MAXN = 1010;//点数的最大值

const int MAXM = 400010;//边数的最大值

const int INF = 0x3f3f3f3f;

struct Edge

{

	int to,next,cap,flow;

}edge[MAXM];//注意是MAXM

int tol;

int head[MAXN];

int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];

void init()

{

	tol = 0;

	memset(head,-1,sizeof(head));

}

//加边，单向图三个参数，双向图四个参数

void addedge(int u,int v,int w,int rw=0)

{

	edge[tol].to = v;edge[tol].cap = w;edge[tol].next = head[u];

	edge[tol].flow = 0;head[u] = tol++;

	edge[tol].to = u;edge[tol].cap = rw;edge[tol].next = head[v];

	edge[tol].flow = 0;head[v]=tol++;

}

//输入参数：起点、终点、点的总数

//点的编号没有影响，只要输入点的总数

int sap(int start,int end,int N)

{

	memset(gap,0,sizeof(gap));

	memset(dep,0,sizeof(dep));

	memcpy(cur,head,sizeof(head));

	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 != -1; 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 != -1; 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 != -1;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 = head[u]; i != -1;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 a[500];

int b[500];

int id[500][500];

bool vis[MAXN];

bool dfs(int u,int pre)

{

	//printf("%d %d\n",pre,u);

	if(vis[u])return true;

	vis[u] = true;

	for(int i = head[u];i != -1;i = edge[i].next)

	{

		int v = edge[i].to;

		if(edge[i].cap <= edge[i].flow)continue;

		if(v == pre)continue;

		if(dfs(v,u))return true;

	}

    vis\[u\] = false;
    return false;

}

int main()

{

    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

    int n,m,K;

	while(scanf("%d%d%d",&n,&m,&K) == 3)

	{

		init();

		int sum1 = 0;

		for(int i = 1;i <= n;i++)

		{

			scanf("%d",&a[i]);

			sum1 += a[i];

			addedge(0,i,a[i]);

		}

		int sum2 = 0;

		for(int i = 1;i <= m;i++)

		{

			scanf("%d",&b[i]);

			sum2 += b[i];

			addedge(i+n,n+m+1,b[i]);

		}

		if(sum1 != sum2)

		{

			printf("Impossible\n");

			continue;

		}

		for(int i = 1;i <= n;i++)

			for(int j = 1;j <= m;j++)

			{

				id[i][j] = tol;

				addedge(i,j+n,K);

			}

		int tmp = sap(0,n+m+1,n+m+2);

		if(tmp != sum1)

		{

			printf("Impossible\n");

			continue;

		}

		memset(vis,false,sizeof(vis));

		bool flag = false;

		for(int i = 1;i <= n;i++)

			if(dfs(i,i))

			{

				flag = true;

				break;

			}

		if(flag)

		{

			printf("Not Unique\n");

			continue;

		}

		printf("Unique\n");

		for(int i = 1;i <= n;i++)

			for(int j = 1;j <= m;j++)

			{

				printf("%d",edge[id[i][j]].flow);

				if(j < m)printf(" ");

				else printf("\n");

			}

	}

    return 0;

}