題目

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:

D [a] [b]
where [a] and [b] are the numbers of two criminals, and they belong to different gangs.

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.

分析與解答

這題就是我在prim算法里提到過的，利用數組的思想，數組可以表示兩個數的關系，那我們就建一個enemy數組，enemy[a]表示a的敵人是b，每次出現一個D的時候，表明a和b是敵人，那b和enemy[a]就是同一個根，連接起來就行

#include <cstdio> #include <cstring> #include <algorithm> #define maxn 130000 using namespace std;int per[maxn], num[maxn], n, m;int find(int x){if(x == per[x])return x;return per[x] = find(per[x]); }void join(int x, int y){int fx = find(x);int fy = find(y);if(fx != fy){per[fx] = fy;//把fy規定為fx祖宗 }return ; } int a,b,ch,count2=0,enemy[100100]; int main (){int t;scanf("%d",&t);while(t--){memset(enemy,0,sizeof(enemy));scanf("%d%d",&n,&m);for(int j=1;j<=n;++j){per[j]=j;}for(int i=0;i<m;++i){scanf(" %c%d%d",&ch,&a,&b);if (ch == 'D') {if (enemy[a]) join(enemy[a],b);if (enemy[b]) join(enemy[b],a);enemy[a] = b;enemy[b] = a;}else {if (find(a) == find(enemy[b])) printf("In different gangs.\n");else if (find(a) == find(b)) printf("In the same gang.\n");else printf("Not sure yet.\n");}}} }

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