poj 1703並查集路徑壓縮
| Language:Default Find them, Catch them
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.) 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. Source 種類並查集加路徑壓縮和食物鏈那道題相似
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#include<iostream> #include<cstdio> #include<cstring> #define maxn 100005 using namespace std; int father[maxn]; int r[maxn]; int rev(int val) { return val; } int real(int v1,int v2) { if(v1==0) return v2; if(v2==0) return v1; if(v1==1&&v2==1) return 0; } int real3(int v1,int v2,int v3) { return real(real(v1,v2),v3); } int find(int x) { if(father[x]==x) return x; int root=find(father[x]); r[x]=real(r[x],r[father[x]]); return father[x]=root; } int unionn(int u,int v,int relation) { int fu=find(u); int fv=find(v); if(fu!=fv) { r[fu]=real3(rev(r[u]),relation,r[v]); father[fu]=fv; return 1; } return 0; } int getreal(int u,int v) { int fu=find(u); int fv=find(v); if(fu!=fv) return -1; return real(r[u],rev(r[v])); } int main() { int t; scanf("%d",&t); while(t--) { int n,k; scanf("%d%d",&n,&k); for(int i=0;i<=n;i++) father[i]=i; memset(r,0,sizeof(r)); while(k--) { char op[10]; int u,v; scanf("%s%d%d",op,&u,&v); if(op[0]=='A') {int relation=getreal(u,v); if(relation==-1) printf("Not sure yet.\n"); else if(relation==0) printf("In the same gang.\n"); else if(relation==1) printf("In different gangs.\n"); } else if(op[0]=='D') unionn(u,v,1); } } return 0; }