hdu 1829 &poj 2492 A Bug's Life（判断二分图、带权并查集）

A Bug's Life

Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 8528 Accepted Submission(s): 2745



Problem Description

Background

Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the opposite gender. In his experiment, individual bugs and their interactions were easy
to identify, because numbers were printed on their backs.

Problem

Given a list of bug interactions, decide whether the experiment supports his assumption of two genders with no homosexual bugs or if it contains some bug interactions that falsify it.

Input

The first line of the input contains the number of scenarios. Each scenario starts with one line giving the number of bugs (at least one, and up to 2000) and the number of interactions (up to 1000000) separated by a single space. In the following lines, each
interaction is given in the form of two distinct bug numbers separated by a single space. Bugs are numbered consecutively starting from one.

Output

The output for every scenario is a line containing "Scenario #i:", where i is the number of the scenario starting at 1, followed by one line saying either "No suspicious bugs found!" if the experiment is consistent with his assumption about the bugs' sexual
behavior, or "Suspicious bugs found!" if Professor Hopper's assumption is definitely wrong.

Sample Input

2
3 3
1 2
2 3
1 3
4 2
1 2
3 4

Sample Output

Scenario #1:
Suspicious bugs found!

Scenario #2:
No suspicious bugs found!

HintHuge input,scanf is recommended.

利用黑白染色，判断是否是二分图。

#include"stdio.h"
#include"string.h"
#include"queue"
using namespace std;
#define N 2005
#define M 1000005
int n,m,t,head
;
int color
,flag;
struct node
{
int u,v,next;
}map[2*M];
void add(int u,int v)
{
map[t].u=u;
map[t].v=v;
map[t].next=head[u];
head[u]=t++;
map[t].u=v;
map[t].v=u;
map[t].next=head[v];
head[v]=t++;
}
void find(int u)
{
int i,v;
for(i=head[u];i!=-1;i=map[i].next)
{
v=map[i].v;
if(color[v]==-1)
{
color[v]=color[u]^1;
find(v);
}
else if(color[v]==color[u])
{
flag=1;
return ;
}
}
return ;
}
int main()
{
int i,T,u,v,cnt=1;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
t=0;
memset(head,-1,sizeof(head));
while(m--)
{
scanf("%d%d",&u,&v);
add(u,v);
}
memset(color,-1,sizeof(color));
flag=0;
for(i=1;i<=n;i++)
{
if(color[i]==-1)
{
color[i]=0;
find(i);
if(flag)
break;
}
}
printf("Scenario #%d:\n",cnt++);
if(flag)
printf("Suspicious bugs found!\n");
else
printf("No suspicious bugs found!\n");
puts("");
}
return 0;
}

带权并查集：

#include"stdio.h"
#include"string.h"
#include"queue"
#include"vector"
#include"stack"
#include"algorithm"
using namespace std;
#define N 2005
#define min(a,b) (a<b?a:b)
int pre
,gen
;
int find(int k)
{
if(k==pre[k])
return k;
int t=find(pre[k]);     //不能立即更新父节点与根节点相同
gen[k]=gen[k]^gen[pre[k]];//因为该节点的性别和它的父节点相反
return pre[k]=t;  //确定该点性别之后才能把该点父节点更新为根节点
}
int Union(int x,int y)
{
int a,b;
a=find(x);
b=find(y);
if(a==b)       //x,y的根节点相同
{
if(gen[x]==gen[y]) //判断他们是否同性别
return 1;
return 0;
}
pre[a]=b;
gen[a]=(gen[x]+gen[y]+1)&1;
return 0;
}
int main()
{
int i,u,v,n,cnt=1,T,m;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
for(i=0;i<=n;i++)
pre[i]=i;
memset(gen,0,sizeof(gen));
int flag=0;
while(m--)
{
scanf("%d%d",&u,&v);
if(flag)
continue;
flag=Union(u,v);
}
printf("Scenario #%d:\n",cnt++);
if(flag)
printf("Suspicious bugs found!\n");
else
printf("No suspicious bugs found!\n");
puts("");
}
return 0;
}