码迷,mamicode.com
首页 > Windows程序 > 详细

hdu 4888 Redraw Beautiful Drawings(最大流,判环)

时间:2017-05-15 09:50:28      阅读:247      评论:0      收藏:0      [点我收藏+]

标签:pid   str   矩阵   ons   init   ems   ring   else   content   

http://acm.hdu.edu.cn/showproblem.php?pid=4888


加入一个源点与汇点,建图例如以下

1. 源点 -> 每一行相应的点,流量限制为该行的和

2. 每一行相应的点 -> 每一列相应的点,流量限制为 K

3. 每一列相应的点 -> 汇点,流量限制为该列的和

求一遍最大流,若最大流与矩阵之和相等,说明有解,否则无解。推断唯一解,是推断残量网络中是否存在一个长度大于2的环。若存在说明有多解,否则有唯一解,解就是每条边行i->列j的流量。


#include <stdio.h>
#include <iostream>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#include <algorithm>
#define LL long long
#define _LL __int64
#define eps 1e-12
#define PI acos(-1.0)
#define C 240
#define S 20
using namespace std;

const int INF = 0x3f3f3f3f;

const int maxn = 810;
const int maxm = 160000+810;

struct node
{
    int u,v,w,next;
} edge[maxm << 1];

int cnt,head[maxn];
int n,m,k,s,t;
int nn[410],mm[410];
int maxflow;
int vis[maxn];
int dis[maxn];

void init()
{
    cnt = 0;
    memset(head,-1,sizeof(head));
}

void add(int u, int v, int f)
{
    edge[cnt] = (struct node){u,v,f,head[u]};
    head[u] = cnt++;

    edge[cnt] = (struct node){v,u,0,head[v]};
    head[v] = cnt++;
}

bool bfs()
{
    queue<int>que;
    while(!que.empty()) que.pop();
    memset(dis,-1,sizeof(dis));
    dis[s] = 0;
    que.push(s);

    while(!que.empty())
    {
        int u = que.front();
        que.pop();

        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].v;
            if(dis[v] == -1 && edge[i].w)
            {
                dis[v] = dis[u]+1;
                que.push(v);
            }
        }
    }
    if(dis[t] == -1)
        return false;
    return true;
}

int dfs(int u, int delta)
{
    if(u == t || delta == 0) return delta;
    int dd;
    int ret = 0;
    for(int i = head[u]; i != -1 && delta; i = edge[i].next)
    {
        if(dis[edge[i].v] == dis[u]+1 && (dd = dfs(edge[i].v,min(delta,edge[i].w))))
        {
            edge[i].w -= dd;
            edge[i^1].w += dd;
            delta -= dd;
            ret += dd;
            if(delta == 0)
                return ret;
        }
    }
	dis[u] = -1;
	return ret;
}

int Dinic()
{
    int ret = 0;
    while(bfs())
    {
        ret += dfs(s,INF);
    }
    return ret;
}

int dfs_1(int u,int fa)
{
    for(int i=head[u]; i!=-1; i=edge[i].next)
    {
        if(i==(fa^1)) continue;
        if(edge[i].w)
        {
            if(vis[edge[i].v]) return 1;
            vis[edge[i].v]=1;
            if(dfs_1(edge[i].v,i)) return 1;
            vis[edge[i].v]=0;
        }
    }
    return 0;
}

void putmat()
{
	int f[410][410];
	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].v;
			if(v > n && v <= n+m)
				f[u][v-n] = k - edge[i].w;
		}
	}
	for(int i = 1; i <= n; i++)
	{
		for(int j = 1; j < m; j++)
			printf("%d ",f[i][j]);
		printf("%d\n",f[i][m]);
	}
}

int main()
{
    while(~scanf("%d %d %d",&n,&m,&k))
    {
    	init();
        s = 0;
        t = n+m+1;
        int sum1 = 0;
        int sum2 = 0;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&nn[i]);
            add(s,i,nn[i]);
            sum1 += nn[i];
			for(int j = 1; j <= m; j++)
				add(i,j+n,k);
        }

        for(int i = 1; i <= m; i++)
        {
            scanf("%d",&mm[i]);
            add(i+n,t,mm[i]);
            sum2 += mm[i];
        }

        if(sum1 != sum2)
        {
            printf("Impossible\n");
        }

		else
		{
			maxflow = Dinic();

			if(maxflow != sum1)
			{
				printf("Impossible\n");
			}
			else
			{
				int flag = 0;
				memset(vis,0,sizeof(vis));
				for(int i = 1; i <= n; i++)
				{
					if(dfs_1(i,-1))
					{
						flag = 1;
						break;
					}
				}
				if(flag == 1)
					printf("Not Unique\n");

				else
					putmat();
			}
		}
    }
    return 0;
}





hdu 4888 Redraw Beautiful Drawings(最大流,判环)

标签:pid   str   矩阵   ons   init   ems   ring   else   content   

原文地址:http://www.cnblogs.com/jzssuanfa/p/6854807.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!