杭电4263——Red/Blue Spanning Tree

Red/Blue Spanning Tree

链接：http://acm.hdu.edu.cn/showproblem.php?pid=4263

Time Limit: 10000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1283 Accepted Submission(s): 514

Problem Description

Given an undirected, unweighted, connected graph, where each edge is colored either blue or red, determine whether a spanning tree with exactly k blue edges exists.

Input

There will be several test cases in the input. Each test case will begin with a line with three integers:
n m k
Where n (2≤n≤1,000) is the number of nodes in the graph, m (limited by the structure of the graph) is the number of edges in the graph, and k (0≤k<n) is the number of blue edges desired in the spanning tree.
Each of the next m lines will contain three elements, describing the edges:
c f t
Where c is a character, either capital ‘R’ or capital ‘B’, indicating the color of the edge, and f and t are integers (1≤f,t≤n, t≠f) indicating the nodes that edge goes from and to. The graph is guaranteed to be connected, and there is guaranteed to be at most one edge between any pair of nodes.
The input will end with a line with three 0s.

Output

For each test case, output single line, containing 1 if it is possible to build a spanning tree with exactly k blue edges, and 0 if it is not possible. Output no extra spaces, and do not separate answers with blank lines.

Sample Input

3 3 2
B 1 2
B 2 3
R 3 1
2 1 1
R 1 2
0 0 0

Sample Output

1
0

思路：两次最小生成树

正确代码：

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
int s[2000];
int n,m,k;
struct node{
	char str[5];//一定要是个字符数组不然过不了 
	int x;
	int y;
	int f;
} sj[1000005];
void init(){
	for(int i=0;i<=n;i++){
		s[i]=i;
	}
}
int find(int x){
	if(s[x]!=x){
		s[x]=find(s[x]);
	}
	return s[x];
}
void merge(int x,int y){
	int x1=find(x);
	int y1=find(y);
	if(x1!=y1){
		s[x1]=y1;
	}
}
bool cmp1(node a,node b){
	return a.f<b.f;
}
bool cmp2(node a,node b){
	return a.f>b.f;
}
int main(){
	/*ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);*/
	//一定要用scanf和printf,不要用cin，cout会超时 
	while(scanf("%d%d%d",&n,&m,&k)!=EOF){ 
		if(!n&&!m&&!k){
			break;
		}
		init();
		int c1=0;
		int c2=0;
		for(int i=0;i<m;i++){
			scanf("%s %d%d",&sj[i].str,&sj[i].x,&sj[i].y);
			//cin>>sj[i].str>>sj[i].x>>sj[i].y;
			if(sj[i].str[0]=='B'){
				sj[i].f=1;
			}
			else{
				sj[i].f=0;
			}
		}
		sort(sj,sj+m,cmp1);//蓝边最少的时候 
		for(int i=0;i<m;i++){
			if(find(sj[i].x)!=find(sj[i].y)){
				merge(sj[i].x,sj[i].y);
				if(sj[i].f==1){
					c1++;
				}
			}
		}
		sort(sj,sj+m,cmp2);//蓝边最多的时候 
		init();
		for(int i=0;i<m;i++){
			if(find(sj[i].x)!=find(sj[i].y)){
				merge(sj[i].x,sj[i].y);
				if(sj[i].f==1){
					c2++;
				}
			}
		}
		if(c1<=k&&k<=c2){ 
			printf("1\n");
		}
		else{
			printf("0\n");
		}	
	}
	return 0;
}