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POJ 1703 Find them, Catch them（并查集拓展）

2015-08-20 20:32 183 查看
Find them, Catch them
Time Limit: 1000MSMemory Limit: 10000K
Total Submissions: 36534Accepted: 11209
DescriptionThe 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] 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. InputThe 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.OutputFor 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.
[b]    题意：在这个城市里有两个黑帮团伙，现在给出N个人，问任意两个人他们是否在同一个团伙  1.输入D x y代表x于y不在一个团伙里  2.输入A x y要输出x与y是否在同一团伙或者不确定他们在同一个团伙里
#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<stdlib.h>using namespace std;const int maxn = 100010;int bin[maxn],family[maxn];int n,m;int findx(int x){    int r;    if(bin[x]!=x){        r = findx(bin[x]);        family[x] = family[x]^family[bin[x]];        return bin[x] = r;    }    return x;}void bing(int x,int y){    int fx = findx(x);    int fy = findx(y);    bin[fx] = fy;    family[fx] = ~(family[y]^family[x]);}int main(){    int T;    scanf("%d",&T);    while(T--){        char str[5];        int x,y;        scanf("%d%d",&n,&m);        for(int i=0;i<=n;i++){            bin[i] = i;            family[i] = 0;        }        for(int i=0;i<m;i++){            scanf("%s%d%d",&str,&x,&y);            if(strcmp(str,"D") == 0){                bing(x,y);            }else{                if(n == 2){                    printf("In different gangs.\n");                }else if(findx(x) == findx(y)){                    if(family[x] == family[y]){                        printf("In the same gang.\n");                    }else{                        printf("In different gangs.\n");                    }                }else{                    printf("Not sure yet.\n");                }            }        }    }    return 0;}

                                            

                                        

                        
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