"Codevs2597" gang

Description

A group of robbers appeared in Chicago in 1920. If two robbers meet, then they are either friends or enemies. And one thing is for sure, that is:

My friend's friend is my friend;

The enemy of my enemy is also my friend.

Two robbers are the same gang condition is when and only if they are friends. Now give you some information about the robbers and ask you how many bandit gangs there are.

Input

The first line of the input file gangs.in is an integer N (2<=n<=1000) that represents the number of robbers (from 1 to N). The second line, M (1<=m<=5000), indicates the number of information bars about the robber. The following m lines, each line may be f p Q or E P Q (1<=p q<=n), F means P and q are friends, E means P and q are enemies. The input data guarantees that the information is not contradictory.

Output

The output file gangs.out only one row, indicating the maximum number of possible gangs.

Sample Input

6
4
E 1 4
F 3 5
F 4 6
E 1 2

Sample Output

3

HINT

2<=n<=1000

1<=m<=5000

1<=p q<=n

Exercises

And check set, encounter friends Merge, Encounter enemy to add an edge to the diagram, the last for a point, if X and y are connected, x and z, merge Y and Z (enemy is Friends)

#include <iostream>using namespace std;const int n=10010;const int m=50010;struct node{int next,to;} E[m];int head[n],fa[n];int n,m,x,y,cnt=0,ans;char ch;void ins (int u,int v) {e[++cnt].to=v; e[cnt].next=head[u]; Head[u] =cnt;} int root (int x) {if (fa[x]!=x) Fa[x]=root (fa[x]); return fa[x];} void Union (int x,int y) {fa[x]=y;} int main () {cin>>n>>m;for (int i=1;i<=n;i++) fa[i]=i;for (int i=1;i<=m;i++) {cin>>ch>>x >>y;if (ch== ' E ') {ins (x, y); ins (y,x);} Else{if (Root (x)!=root (y)) Union (Fa[x],fa[y]);}} for (int k=1;k<=n;k++) for (int i=head[k];i;i=e[i].next) for (int j=head[e[i].to];j;j=e[j].next) if (k!=e[j].to& &root (k)!=root (e[j].to)) Union (fa[k],fa[e[j].to]); for (int i=1;i<=n;i++) if (fa[i]==i) Ans++;cout<<ans;}

　　

"Codevs2597" gang