HDU 4263 Red/Blue Spanning Tree【最小生成树原理】

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

1
0

题目大意： 给出m条边，是否有满足K条蓝色形成的最小生成数；
思路：根据最小生成树原理；你先把红边放在前面，贪心，求出一个用了k1条蓝边的生成树；再把蓝边放在前面，贪心，求出一个用了k2条蓝边的生成树
　　　　如果满足k1<=k<=k2，则输出1；否则输出0；

代码入下：

View Code

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int f[10500];
struct point
{
int ch, st, end;
}p[1100005];
int find(int x)
{
return x==f[x]?x:f[x]=find(f[x]);
}
int cmp1(point x, point y)
{
return x.ch<y.ch;
}
int cmp2(point x, point y)
{
return x.ch>y.ch;
}
int main()
{
int m, n, k, i, j, ans1, ans2;
while(scanf("%d%d%d", &n, &m, &k)!=EOF)
{
if(n==0&&m==0&&k==0)
break;
//    point p[1100005];  定义在main()内脑袋短路了,,,RE 了N次
for(i=0; i<m; i++)
{
getchar();
scanf("%c%d%d", &p[i].ch, &p[i].st, &p[i].end);
}
sort(p, p+m, cmp1); //蓝的放在前面
for(i=0; i<=n; i++)
f[i]=i;
ans1=0;
for(i=0; i<m; i++)
{
int xx=find(p[i].st), yy=find(p[i].end);
if(xx!=yy)
{
f[xx]=yy;
if(p[i].ch=='B')
ans1++;
}
}
sort(p, p+m, cmp2); //红的放在前面
for(i=0; i<=n; i++)
f[i]=i;
ans2=0;
for(i=0; i<n; i++)
{
int xx=find(p[i].st), yy=find(p[i].end);
if(xx!=yy)
{
f[xx]=yy;
if(p[i].ch=='B')
ans2++;
}
}
if(k>=ans2&&k<=ans1)
printf("1\n");
else
printf("0\n");
}
}