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最大流判断多解

建图:

源点连接到每一个代表行的节点容量为行总和,每一个代表列的节点连接到汇点容量为列总和,行和列之间互相连接容量为Limit

多解:

做一遍ISAP后,在残量图上DFS看能否找到点数大于2的环即可

Redraw Beautiful Drawings

2 2 4
4 2
4 2  
4 2 2
2 2 5 0
5 4
1 4 3
9
1 2 3 3

Not Unique
Impossible
Unique
1 2 3 3

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=1000;
const int maxm=1001000;
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,limit;
int a[maxn],b[maxn];
int id[maxn][maxn];

bool vis[maxn];

bool dfs(int u,int pre)
{
	if(vis[u]) return true;
	vis[u]=true;
	for(int i=Adj[u];~i;i=edge[i].next)
	{
		int v=edge[i].to;
		if(edge[i].flow>=edge[i].cap) continue;
		if(v==pre) continue;
		if(dfs(v,u)) return true;
	}
	vis[u]=false;
	return false;
}

int main()
{
	while(scanf("%d%d%d",&n,&m,&limit)!=EOF)
	{
		init();
		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)
		{
			puts("Impossible");
			continue;
		}

		/**************build graph *****************/
		for(int i=1;i<=n;i++)
			for(int j=1;j<=m;j++)
			{
				id[i][j]=Size;
				addedge(i,n+j,limit);
			}
		for(int i=1;i<=n;i++)
			addedge(0,i,a[i]);
		for(int i=1;i<=m;i++)
			addedge(n+i,n+m+1,b[i]);
		/**************build graph *****************/

		int MaxFlow=sap(0,n+m+1,n+m+1+2);
		//cout<<"MaxFlow: "<<MaxFlow<<endl;
		if(MaxFlow!=sum1)	
		{
			puts("Impossible");
			continue;
		}
		else
		{
			memset(vis,0,sizeof(vis));
			bool flag=false;
			for(int i=1;i<=n&&!flag;i++)
				if(dfs(i,i)) flag=true;
			if(flag)
			{
				puts("Not Unique");
				continue;
			}
			puts("Unique");
			for(int i=1;i<=n;i++)
				for(int j=1;j<=m;j++)
				{
					printf("%d%c",edge[id[i][j]].flow,(j==m)?10:32);
				}
		}	
	}
	return 0;
}

HDOJ 4888 Redraw Beautiful Drawings

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原文地址：http://blog.csdn.net/ck_boss/article/details/39942313