Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16075 Accepted Submission(s): 6677
3 3 1 2 1 1 3 2 2 3 4 1 3 2 3 2 0 100
3 ?
这是一道带权值的并查集,因为是中文题, 题意也就不用多说了。。
所以我是用结构体写的。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<sstream>
#include<cmath>
using namespace std;
#define f1(i, n) for(int i=0; i<n; i++)
#define f2(i, n) for(int i=1; i<=n; i++)
#define f3(i, n) for(int i=n; i>=1; i--)
#define f4(i, n) for(int i=1; i<n; i++)
#define M 10050
int f[M];
int r[M];
int ans;
int t;
int n, m;
int coun;
struct node
{
int x;
int y;
int cost; //花费
}q[M];
int cmp(node x1, node y1)
{
return x1.cost < y1.cost;
}
int find(int x) //并查集的find
{
return f[x] == x ? x:f[x] = find( f[x] );
}
void Kruskal()
{
sort(q, q+n, cmp);
f2(i, n)
{
int xx = find(q[i].x);
int yy = find(q[i].y);
if( xx!=yy ) //当不是统一集合时。。
{
ans+=q[i].cost;
f[yy] = xx;
coun --; //连起来一条路
}
}
}
int main()
{
while(scanf("%d%d", &n, &m) &&n)
{
ans = 0;
t = 0;
coun = m;
f2(i, m) f[i] = i; //初始化
f2(i, n)
scanf("%d%d%d", &q[i].x, &q[i].y, &q[i].cost);
Kruskal();
if(coun==1)
printf("%d\n", ans);
else
printf("?\n");
}
return 0;
}
HDU 1863:畅通工程(带权值的并查集),布布扣,bubuko.com
原文:http://blog.csdn.net/u013487051/article/details/37805331