Red/Blue Spanning Tree
Time Limit: 2000ms
Memory Limit: 131072KB
This problem will be judged on HDU. Original ID: 4263
64-bit integer IO format: %I64d Java class name: Main
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, andk (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
Source
The University of Chicago Invitational Programming Contest 2012
解题：Kruskal最小生成树模型！先把红边放在前面，做一次Kruskal,再把蓝边放在前面，做一次Kruskal，记录两次蓝边的选用情况分布为x,y!if x <= k <= y那么输出1，否则输出0.
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cmath>
5 #include <algorithm>
6 #include <climits>
7 #include <vector>
8 #include <queue>
9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 1010;
18 struct arc{
19 int u,v;
20 char color;
21 arc(int x = 0,int y = 0,char ch = ‘#‘):u(x),v(y),color(ch){}
22 };
23 bool cmp1(const arc &x,const arc &y){
24 return x.color < y.color;
25 }
26 bool cmp2(const arc &x,const arc &y){
27 return x.color > y.color;
28 }
29 arc e[1100010];
30 int n,m,k,uf[maxn];
31 int Find(int x){
32 if(x != uf[x]){
33 uf[x] = Find(uf[x]);
34 }
35 return uf[x];
36 }
37 void kruskal(int &cnt,char ch){
38 if(ch == ‘B‘) sort(e,e+m,cmp1);
39 else sort(e,e+m,cmp2);
40 int i,j,tx,ty;
41 for(i = 0; i <= n; i++) uf[i] = i;
42 for(cnt = i = 0; i < m; i++){
43 tx = Find(e[i].u);
44 ty = Find(e[i].v);
45 if(tx != ty){
46 uf[tx] = ty;
47 if(e[i].color == ‘B‘) cnt++;
48 }
49 }
50 }
51 int main() {
52 int i,j,u,v,x,y;
53 char s[5];
54 while(scanf("%d %d %d",&n,&m,&k),n||m||k){
55 for(i = 0; i < m; i++){
56 scanf("%s %d %d",s,&u,&v);
57 e[i] = arc(u,v,s[0]);
58 }
59 kruskal(x,‘R‘);
60 kruskal(y,‘B‘);
61 if(k >= x && k <= y) puts("1");
62 else puts("0");
63 }
64 return 0;
65 }
比较另类但是比较好写的写法，摘自。。。
1 #include<stdio.h>
2 #include<string.h>
3 #include<iostream>
4 #include<map>
5 #include<queue>
6 #include<algorithm>
7 using namespace std;
8
9 const int maxN = 2100;
10 bool vis[maxN];
11
12 struct node{
13 int x, y;
14 }p[maxN * 100];
15
16 int find(int u,int *f) {
17 if(f[u] == u)
18 return u;
19 return f[u] = find(f[u], f);
20 }
21 bool fun(int u,int v,int *f) {
22 int px = find(u, f), py = find(v, f);
23 if(px != py) {
24 f[px] = py;
25 return true;
26 }
27 return false;
28 }
29
30 int main() {
31 int N, M, K;
32 while(scanf("%d%d%d", &N, &M, &K) && (N + M + K)) {
33 int f1[maxN], f2[maxN], f3[maxN], num = 0;
34 for(int i = 0;i <= N;++ i)
35 f1[i] = f2[i] = f3[i] = i;
36 for(int i = 0;i < M; ++ i) {
37 char s[10];
38 int u, v;
39 scanf("%s%d%d", s, &u, &v);
40 if(s[0] == ‘R‘)
41 fun(u, v, f2);
42 else {
43 p[num].x = u;
44 p[num ++].y = v;
45 }
46 }
47 memset(vis, 0, sizeof(vis));
48 int sum = 0, ans = 0;
49 for(int i = 1;i <= N; ++ i) {
50 int px = find(i, f2);
51 if(!vis[px]) {
52 sum ++;
53 vis[px] = true;
54 }
55 }
56 for(int i = 0;i < num; ++ i)
57 if(fun(p[i].x, p[i].y, f2)) {
58 sum --;
59 K --;
60 fun(p[i].x, p[i].y, f3);
61 p[i].x = p[i].y = -1;
62 }
63 int flag = 0;
64 if(sum == 1 && K >= 0) {
65 for(int i = 0;i < num && K > 0; ++ i)
66 if(p[i].x > 0 && fun(p[i].x, p[i].y, f3)) {
67 K --;
68 fun(p[i].x, p[i].y, f2);
69 }
70 }
71 if(sum == 1 && K == 0)flag = 1;
72 printf("%d\n", flag);
73 }
74 }
BNUOJ 26229 Red/Blue Spanning Tree,布布扣,bubuko.com