和HDOJ4888是一样的问题,最大流推断多解

1.把ISAP卡的根本出不来结果,仅仅能把全为0或者全为满流的给特判掉......

2.在残量网络中找大于2的圈要用一种类似tarjian的方法从汇点開始找,推断哪些点没有到汇点

A simple Gaussian elimination problem.

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1170    Accepted Submission(s): 377

Problem Description
Dragon is studying math. One day, he drew a table with several rows and columns, randomly wrote numbers on each elements of the table. Then he counted the sum of each row and column. Since he thought the map will be useless after he got the sums, he destroyed
the table after that.



However Dragon's mom came back and found what he had done. She would give dragon a feast if Dragon could reconstruct the table, otherwise keep Dragon hungry. Dragon is so young and so simple so that the original numbers in the table are one-digit number (e.g.
0-9).



Could you help Dragon to do that?
 
Input
The first line of input contains only one integer, T(<=30), the number of test cases. Following T blocks, each block describes one test case.



There are three lines for each block. The first line contains two integers N(<=500) and M(<=500), showing the number of rows and columns.



The second line contains N integer show the sum of each row.



The third line contains M integer show the sum of each column.
 
Output
Each output should occupy one line. Each line should start with "Case #i: ", with i implying the case number. For each case, if we cannot get the original table, just output: "So naive!", else if we can reconstruct the table by more than one ways, you should
output one line contains only: "So young!", otherwise (only one way to reconstruct the table) you should output: "So simple!".
 
Sample Input
3
1 1
5
5
2 2
0 10
0 10
2 2
2 2
2 2
 
Sample Output
Case #1: So simple!
Case #2: So naive!
Case #3: So young!
 
Author
BJTU
 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm> #pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; const int maxn=20000;
const int maxm=500000;
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;
int a[maxn],b[maxn]; bool vis[maxn],no[maxn];
int Stack[maxm],stk; bool dfs(int u,int pre,bool flag)
{
vis[u]=true;
Stack[stk++]=u;
for(int i=Adj[u];~i;i=edge[i].next)
{
int v=edge[i].to;
if(v==pre) continue;
if(edge[i].flow>=edge[i].cap) continue;
if(!vis[v])
{
if(dfs(v,u,edge[i^1].cap>edge[i^1].flow)) return true;
}
else if(!no[v]) return true;
}
if(flag==false)
{
while(true)
{
int v=Stack[--stk];
no[v]=true;
if(v==u) break;
}
}
return false;
} int main()
{
int T_T,cas=1;
scanf("%d",&T_T);
while(T_T--)
{
scanf("%d%d",&n,&m);
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)
{
printf("Case #%d: So naive!\n",cas++);
continue;
}
if(sum1==sum2&&((sum1==0)||(sum1==n*m*9)))
{
printf("Case #%d: So simple!\n",cas++);
continue;
} /*************build graph*****************/
init();
for(int i=1;i<=n;i++) addedge(0,i,a[i]);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
addedge(i,n+j,9);
for(int i=1;i<=m;i++) addedge(i+n,n+m+1,b[i]);
/*************build graph*****************/
int MaxFlow=sap(0,n+m+1,n+m+2); if(MaxFlow!=sum1)
{
printf("Case #%d: So naive!\n",cas++);
continue;
}
stk=0;
memset(vis,0,sizeof(vis));
memset(no,0,sizeof(no));
if(dfs(n+m+1,n+m+1,0))
{
printf("Case #%d: So young!\n",cas++);
}
else
{
printf("Case #%d: So simple!\n",cas++);
}
}
return 0;
}

HDOJ 4975 A simple Gaussian elimination problem.的更多相关文章

  1. HDU 4975 A simple Gaussian elimination problem.

    A simple Gaussian elimination problem. Time Limit: 1000ms Memory Limit: 65536KB This problem will be ...

  2. hdu 4975 A simple Gaussian elimination problem.(网络流,推断矩阵是否存在)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4975 Problem Description Dragon is studying math. One ...

  3. hdu - 4975 - A simple Gaussian elimination problem.(最大流量)

    意甲冠军:要在N好M行和列以及列的数字矩阵和,每个元件的尺寸不超过9,询问是否有这样的矩阵,是独一无二的N(1 ≤ N ≤ 500) , M(1 ≤ M ≤ 500). 主题链接:http://acm ...

  4. hdu 4975 A simple Gaussian elimination problem 最大流+找环

    原题链接 http://acm.hdu.edu.cn/showproblem.php?pid=4975 这是一道很裸的最大流,将每个点(i,j)看作是从Ri向Cj的一条容量为9的边,从源点除法连接每个 ...

  5. hdu4975 A simple Gaussian elimination problem.(正确解法 最大流+删边判环)(Updated 2014-10-16)

    这题标程是错的,网上很多题解也是错的. http://acm.hdu.edu.cn/showproblem.php?pid=4975 2014 Multi-University Training Co ...

  6. A simple Gaussian elimination problem.(hdu4975)网络流+最大流

    A simple Gaussian elimination problem. Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65 ...

  7. A simple Gaussian elimination problem.

    hdu4975:http://acm.hdu.edu.cn/showproblem.php?pid=4975 题意:给你一个n*m的矩阵,矩阵中的元素都是0--9,现在给你这个矩阵的每一行和每一列的和 ...

  8. hdu4975 A simple Gaussian elimination problem.(最大流+判环)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4975 题意:和hdu4888基本一样( http://www.cnblogs.com/a-clown/ ...

  9. BNU 4356 ——A Simple But Difficult Problem——————【快速幂、模运算】

    A Simple But Difficult Problem Time Limit: 5000ms Memory Limit: 65536KB 64-bit integer IO format: %l ...

随机推荐

  1. 2015 Multi-University Training Contest 7 hdu 5372 Segment Game

    Segment Game Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Tota ...

  2. ASP.NET-表单验证-DataAnnotations

    DataAnnotations  [数据注解,数据注释] 需要引入两个脚本文件 <script src="@Url.Content("~/Scripts/jquery.val ...

  3. java properties类读取配置文件

    1.JAVA Properties类,在java.util包里,具体类是java.util.properties.Properties类继承自Hashtable类并且实现了Map接口,也是使用一种键值 ...

  4. linux gnome kde点滴

    2014.12.08 下面切换的方法对于fedora 17没有效果,对于fedora 17, 要使用system-switch-displaymanager,出现 点击相应的选项,然后就进入相应的启动 ...

  5. Android学习之GridView图片布局适配经验

    開始解说这篇博客之前,我想问一下,当布局相似GridView这样的多列布局时,我们该怎么布局,才干更好的去适配呢? 扣张图来展示一下 比如这样的需求,三张图片均分屏幕 实现方法: 1.切图固定,比如是 ...

  6. 安全风控的CAP原理和BASE思想

    CAP原理最多实现两个,需要牺牲一个来满足其他两个:

  7. Linux xhost命令详解

    xhost 命令用途 控制什么人可以访问当前主机上的增强 X-Windows. 语法 xhost [ + | - ] [ Name ] "+"表示增加,"-"表 ...

  8. 14.boost最小生成树 kruskal_min_spainning_tree

    #include <iostream> #include <boost/config.hpp> //图(矩阵实现) #include <boost/graph/adjac ...

  9. 使用Swing组件编写一个支持中文文本编辑程序ChineseTextEdit.java

      import javax.swing.*; import java.awt.*; import java.awt.event.*; import java.io.*; public class C ...

  10. BZOJ 3160 FFT+Manacher

    思路: 这道题思路好奇怪--. 我们先要知道关于x (x可以是间隙) 对称的有几对字母 显然暴力是n^2的 那怎么办呢 先把所有'a'看成1 'b'看成0 意外的发现 这不就是卷积嘛 再倒过来搞一搞 ...