poj1703 Find them, Catch them(并查集的应用)
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 32225 | Accepted: 9947 |
Description
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
Output
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. 题目大意:N 表示共有 N 个帮派;下面有 M 条个询问
有两个帮派,现在告诉你一些互相敌对的点对,然后让你判断某两个点之间的关系,并首先判两个点关系是否确定;
用并查集的思想只要两个点在一个集合里就可以了,直接用并查集然后时候在同一个集合;
AC代码:
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
// 0 代表木有关系,1表示不同类,2代表同类
const int N = 1e5+;
int m;
int f[N],c[N];
int init()
{
for(int i=; i<=m; i++)
{
f[i] = i;
c[i] = ;
}
}
int xfind(int x)
{
if(f[x] == x)
return x;
int t = f[x];
f[x] =xfind(f[x]);
c[x] = (c[x]+c[t])%;
return f[x];
}
int solve(char ch,int x,int y )
{
int fx = xfind(x);
int fy = xfind(y);
if(fx != fy)
{
if(ch == 'A') return ; //返回值0,表示不确定
f[fx] =fy;
if((c[x]+c[y])% ==)
c[fx] = ;
}
else
{
if(ch == 'A')
{
if(c[x] != c[y]) return ; //返回值1,表示在不同的帮派
if(c[x] == c[y]) return ; //返回值2,表示在相同的帮派
}
}
if(ch == 'D') return ;//这种情况不同在考虑的
}
int main( )
{
int T,n,x,y;
char s;
scanf("%d",&T);
while(T--)
{
scanf("%d %d",&m,&n);
init(); //初始化并查集
for(int i=; i<n; i++)
{
scanf(" %c %d %d",&s,&x,&y);
int flag = solve(s,x,y);
if(flag == ) printf("Not sure yet.\n");
else if(flag == ) printf("In different gangs.\n");
else if(flag == ) printf("In the same gang.\n");
}
}
return ;
}
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