Find them, Catch them
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 37242 | Accepted: 11483 |
Description
The police office in Tadu City decides to say ends to the chaos, as launch actions to root up the TWO gangs in the city, Gang Dragon and Gang Snake. However, the police first needs to identify which gang a criminal belongs to. The present question is, given
two criminals; do they belong to a same clan? You must give your judgment based on incomplete information. (Since the gangsters are always acting secretly.)
Assume N (N <= 10^5) criminals are currently in Tadu City, numbered from 1 to N. And of course, at least one of them belongs to Gang Dragon, and the same for Gang Snake. You will be given M (M <= 10^5) messages in sequence, which are in the following two kinds:
1. D [a] [b]
where [a] and [b] are the numbers of two criminals, and they belong to different gangs.
2. A [a] [b]
where [a] and [b] are the numbers of two criminals. This requires you to decide whether a and b belong to a same gang.
Assume N (N <= 10^5) criminals are currently in Tadu City, numbered from 1 to N. And of course, at least one of them belongs to Gang Dragon, and the same for Gang Snake. You will be given M (M <= 10^5) messages in sequence, which are in the following two kinds:
1. D [a] [b]
where [a] and [b] are the numbers of two criminals, and they belong to different gangs.
2. A [a] [b]
where [a] and [b] are the numbers of two criminals. This requires you to decide whether a and b belong to a same gang.
Input
The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. Each test case begins with a line with two integers N and M, followed by M lines each containing one message as described above.
Output
For each message "A [a] [b]" in each case, your program should give the judgment based on the information got before. The answers might be one of "In the same gang.", "In different gangs." and "Not sure yet."
Sample Input
1 5 5 A 1 2 D 1 2 A 1 2 D 2 4 A 1 4
Sample Output
Not sure yet. In different gangs. In the same gang.
题意是在Tadu这个城市有两个团伙。
给出D i j的意思是说i和j不是一个团伙的。
给出A i j的意思是根据已知的信息，能不能确定i和j是一个团伙的，如果确定是一个团伙的，输出“In different gangs.”。如果确定不是一个团伙的，输出“In the same gang.”。如果不能确定，输出“Not sure yet.”
之前使用并查集都是判断两个人是不是在一个集合里，这次要确定地判断两个人不在集合里，所以想法就是添加无用的N个元素作为桥梁。i k不在一个集合中，k j不在一个集合中，说明i j在一个集合中，那我把i k j都放入集合中，只不过k是我查询永远都不会用到的元素，所以就是添加元素时 i k+N，k+N j添加到集合中。所以当我查询k+N，j 发现他们在一个集合中时，就恰恰说明了k和j在两个团伙里面。
代码：
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;
char oper[5];
int n,m,num;
int pre[200015];
int findpre(int x)
{
while(x!=pre[x])
{
x=pre[x];
}
return x;
}
void union_set(int x,int y)
{
int pre_x=findpre(x);
int pre_y=findpre(y);
if(pre_x == pre_y)
return;
else if(pre_x>pre_y)
{
int temp = pre_x;
pre_x = pre_y;
pre_y = temp;
}
pre[pre_y]=pre_x;
}
bool same(int x,int y)
{
return findpre(x) == findpre(y);
}
int main()
{
//freopen("i.txt","r",stdin);
//freopen("o.txt","w",stdout);
int test,i,temp1,temp2;
scanf("%d",&test);
while(test--)
{
scanf("%d%d",&n,&m);
for(i=1;i<=2*n;i++)
{
pre[i]=i;
}
for(i=1;i<=m;i++)
{
scanf("%s%d%d",oper,&temp1,&temp2);
if(oper[0]=='A')
{
if(same(temp1,temp2))
{
printf("In the same gang.
");
}
else if(same(temp1+n,temp2)||same(temp1,temp2+n))
{
printf("In different gangs.
");
}
else
{
printf("Not sure yet.
");
}
}
else if(oper[0]=='D')
{
union_set(temp1,temp2+n);
union_set(temp1+n,temp2);
}
}
}
//system("pause");
return 0;
}POJ2942与此类似，代码：
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;
int n,m;
int pre[200015];
int findpre(int x)
{
while(x!=pre[x])
{
x=pre[x];
}
return x;
}
void union_set(int x,int y)
{
int pre_x=findpre(x);
int pre_y=findpre(y);
if(pre_x == pre_y)
return;
else if(pre_x>pre_y)
{
int temp = pre_x;
pre_x = pre_y;
pre_y = temp;
}
pre[pre_y]=pre_x;
}
bool same(int x,int y)
{
return findpre(x) == findpre(y);
}
int main()
{
//freopen("i.txt","r",stdin);
//freopen("o.txt","w",stdout);
int test,i,j,temp1,temp2;
scanf("%d",&test);
for(i=1;i<=test;i++)
{
printf("Scenario #%d:
",i);
scanf("%d%d",&n,&m);
for(j=1;j<=2*n;j++)
{
pre[j]=j;
}
bool flag=false;
for(j=1;j<=m;j++)
{
scanf("%d%d",&temp1,&temp2);
if(flag)continue;
if(same(temp1,temp2))
{
flag=true;
continue;
}
union_set(temp1,temp2+n);
union_set(temp1+n,temp2);
}
if(flag)
{
printf("Suspicious bugs found!
");
}
else
{
printf("No suspicious bugs found!
");
}
}
//system("pause");
return 0;
}POJ1182是把集合弄成了3个，那与此同理，就是x,x+n,x+2*n的区别，用x,y+n表示A吃B的集合。再对每一个语句进行排除，得到答案。
代码：
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;
int n,m,num;
int pre[150015];
int findpre(int x)
{
while(x!=pre[x])
{
x=pre[x];
}
return x;
}
void union_set(int x,int y)
{
int pre_x=findpre(x);
int pre_y=findpre(y);
if(pre_x == pre_y)
return;
else if(pre_x>pre_y)
{
int temp = pre_x;
pre_x = pre_y;
pre_y = temp;
}
pre[pre_y]=pre_x;
}
bool same(int x,int y)
{
return findpre(x) == findpre(y);
}
int main()
{
//freopen("i.txt","r",stdin);
//freopen("o.txt","w",stdout);
int oper,i,x,y,ans;
scanf("%d%d",&n,&m);
ans=0;
for(i=1;i<=3*n;i++)
{
pre[i]=i;
}
for(i=1;i<=m;i++)
{
scanf("%d%d%d",&oper,&x,&y);
if(x<=0||x>n||y<=0||y>n)
{
ans++;
continue;
}
if(oper==1)
{
if(same(x,y+n)||same(x,y+2*n)||same(x+n,y))
{
ans++;
continue;
}
union_set(x,y);
union_set(x+n,y+n);
union_set(x+2*n,y+2*n);
}
else if(oper==2)
{
if(x==y||same(x,y)||same(x+n,y)||same(x,y+2*n))
{
ans++;
continue;
}
union_set(x,y+n);
union_set(x+n,y+2*n);
union_set(x+2*n,y);
}
}
printf("%d
",ans);
//system("pause");
return 0;
}版权声明：本文为博主原创文章，未经博主允许不得转载。