学步园

思路：把红边设为1，蓝边设为2，求一次红色优先的最小生成树，得出最少的蓝边数，再蓝色优先求一次最小生成树，得出最多的蓝边数，最后只要K在这个范围内就可以

code：

/*
    ID:yueqi
    LANG:C++
    TASK:humble
*/
#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <limits.h>
#include <vector>
#include <bitset>
#include <string>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <string.h>
#include <iostream>
#include <algorithm>
#define Si set<int>
#define LL long long
#define pb push_back
#define PS printf(" ")
#define Vi vector<int>
#define LN printf("\n")
#define lson l,m,rt << 1
#define rson m+1,r,rt<<1|1
#define SD(a) scanf("%d",&a)
#define PD(a) printf("%d",a)
#define SET(a,b) memset(a,b,sizeof(a))
#define FF(i,a) for(int i(0);i<(a);i++)
#define FD(i,a) for(int i(a);i>=(1);i--)
#define FOR(i,a,b) for(int i(a);i<=(b);i++)
#define FOD(i,a,b) for(int i(a);i>=(b);i--)
#define readf freopen("humble.in","r",stdin)
#define writef freopen("humble.out","w",stdout)
const int maxn = 1001;
const long long BigP=999983;
const long long  INF = 0x5fffffff;
const int dx[]={-1,0,1,0};
const int dy[]={0,1,0,-1};
const double pi = acos(-1.0);
const double eps= 1e-7;
using namespace std;
int N,M,K,father[maxn],rank[maxn];
struct edge{
    int u,v;
    int col;//col==1 red,col==2 blue;
}Edge[maxn*maxn];
bool cmp1(const edge a,const edge b){
    return a.col<b.col;
}
bool cmp2(const edge a,const edge b){
    return a.col>b.col;
}
void initUnion(){
    FOR(i,1,N){
        father[i]=i;
        rank[i]=0;
    }
}
int find(int k){
    return k==father[k] ? k:father[k]=find(father[k]);
}
void Union(int x,int y){
    if(rank[x]<rank[y]){
        father[x]=y;
    }else{
        father[y]=x;
        if(rank[x]==rank[y]) rank[x]++;
    }
}
int kruskal(){
    int blue_cnt=0;
    int cnt=0;
    FOR(i,1,M){
        int x=find(Edge[i].u);
        int y=find(Edge[i].v);
        if(x!=y){
            Union(x,y);
            cnt++;
            if(Edge[i].col==2)
                blue_cnt++;
            if(cnt==N-1) break;
        }
    }
    return blue_cnt;
}
int main(){
    while(~scanf("%d%d%d",&N,&M,&K)&&(N+M+K)){
        initUnion();
        char buf[2];
        int u,v;
        FOR(i,1,M){
            scanf("%s",buf);
            scanf("%d%d",&u,&v);
            Edge[i].u=u;
            Edge[i].v=v;
            if(buf[0]=='R') Edge[i].col=1;
            else Edge[i].col=2;
        }
        //最小蓝色生成树
        sort(Edge+1,Edge+M+1,cmp1);
        int minn=kruskal();
        initUnion();
        //最小红色
        sort(Edge+1,Edge+M+1,cmp2);
        int maxx=kruskal();
        if(K>=minn&&K<=maxx) puts("1");
        else puts("0");
    }
    return 0;
}