HDoj-1879-畅通project-并查集

继续畅通project

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 14309    Accepted Submission(s): 6228

Problem Description

省政府“畅通project”的目标是使全省不论什么两个村庄间都能够实现公路交通（但不一定有直接的公路相连，仅仅要能间接通过公路可达就可以）。现得到城镇道路统计表，表中列出了随意两城镇间修建道路的费用，以及该道路是否已经修通的状态。现请你编敲代码，计算出全省畅通须要的最低成本。

 

Input

測试输入包括若干測试用例。每一个測试用例的第1行给出村庄数目N ( 1< N < 100 )；随后的 N(N-1)/2 行相应村庄间道路的成本及修建状态，每行给4个正整数，各自是两个村庄的编号（从1编号到N），此两村庄间道路的成本，以及修建状态：1表示已建，0表示未建。



当N为0时输入结束。

 

Output

每一个測试用例的输出占一行，输出全省畅通须要的最低成本。

 

Sample Input

3
1 2 1 0
1 3 2 0
2 3 4 0
3
1 2 1 0
1 3 2 0
2 3 4 1
3
1 2 1 0
1 3 2 1
2 3 4 1
0

 

Sample Output

3
1
0



#include <stdio.h>
#include<string.h>
#include <algorithm>
using namespace std;
typedef struct Road
{
    int c1, c2, cost, state;
}Road;

bool myCompare(const Road &a, const Road &b)
{
    if(a.cost < b.cost)
        return 1;
    return 0;
}

Road road[5051];
int city[101];

int Find(int n)
{
    if(city[n] == -1)
        return n;
    return city[n] = Find(city[n]);
}

bool Merge(int s1, int s2)
{
    int r1 = Find(s1);
    int r2 = Find(s2);
    if(r1 == r2)
        return 0;
    if(r1 < r2)
        city[r2] = r1;
    else
        city[r1] = r2;
    return 1;
}

int main()
{
    //freopen("input.txt", "r", stdin);
    int n;
    while(scanf("%d", &n) && n)
    {
        int m = n*(n-1)/2;
        memset(city, -1, sizeof(city));
        int count = 0;
        for(int i=0; i<m; ++i)
        {
            scanf("%d %d %d %d", &road[i].c1, &road[i].c2, &road[i].cost, &road[i].state);
            if(road[i].state == 1)
            {
                count ++;
                Merge(road[i].c1, road[i].c2);
            }
        }
        sort(road, road+m, myCompare);
        int sum = 0;
        for(int i=0; i<m; ++i)
        {
            if(Merge(road[i].c1, road[i].c2) && road[i].state == 0)
            {
                count ++;
                sum += road[i].cost;
            }
            if(count == n-1)
                break;
        }
        printf("%d\n", sum);
    }
    return 0;
}

#include <stdio.h>
#include<string.h>
#include <algorithm>
int city[5000];
using namespace std;
struct Road
{
    int c1;
	int c2;
	int cost;
	int state;
};
Road road[110];

bool cmp(Road a,Road b)
{
	if(a.cost<b.cost)
	return 1;
return 0;
}
int find(int a)
{
    if(city[a]==-1)
        return a;
return city[a]=find(city[a]);
}
bool merge(int x, int y)
{
	int fx,fy;
     fx = find(x);
     fy = find(y);
    if(fx == fy)   return 0;
    else if(fx < fy)                //不能省去
           city[fy] = fx;
    else
           city[fx] = fy;
return 1;
} 

int main()
{
    int n, m;
    while(scanf("%d",&n),n)
    {
    	m=n*(n-1)/2;
    	int i;
    	int count = 0;
    	int sum = 0;
        memset(city,-1,sizeof(city));

        for(i=0; i<m; i++)
        {
            scanf("%d %d %d %d",&road[i].c1, &road[i].c2, &road[i].cost, &road[i].state);
            if(road[i].state == 1)
            {
                count ++;
                merge(road[i].c1, road[i].c2);
            }
        }
        sort(road, road+m,cmp);

        for(i=0; i<m; i++)
        {
            if(merge(road[i].c1, road[i].c2) && road[i].state==0)
            {
                count ++;
                sum += road[i].cost;
            }
            if(count == n-1)
                break;
        }
            printf("%d\n",sum);
    }
    return 0;
}