HDU ACM 1879 继续畅通工程

继续畅通工程

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10206    Accepted Submission(s): 4451

Problem Description

省政府“畅通工程”的目标是使全省任何两个村庄间都可以实现公路交通（但不一定有直接的公路相连，只要能间接通过公路可达即可）。现得到城镇道路统计表，表中列出了任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。现请你编写程序，计算出全省畅通需要的最低成本。

 

Input

测试输入包含若干测试用例。每个测试用例的第1行给出村庄数目N ( 1< N < 100 )；随后的 N(N-1)/2 行对应村庄间道路的成本及修建状态，每行给4个正整数，分别是两个村庄的编号（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

每个测试用例的输出占一行，输出全省畅通需要的最低成本。

 

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

 

Author

ZJU

 

Source

浙大计算机研究生复试上机考试-2008年

 

#include <cstdio>
#include <queue>
#include <cstring>

using namespace std;

const int NV = 102;
const int NE = 10000;
const int INF = 1<<30;
int ne, nv, term, tot;
bool vis[NV];
int dist[NV];
struct Edge{
    int v, cost, next;
    Edge(){}
    Edge(int a, int c){v = a, cost = c;}
    Edge(int a, int c, int d){v = a, cost = c, next = d;}
    bool operator < (const Edge& x) const
    {
        return cost > x.cost; 
    }
}edge[NE];
int eh[NV];

int prim(int s = 1)
{
    for(int i = 1; i <= nv; ++i) dist[i] = i == s ? 0 : INF;
    priority_queue<Edge> que;
    que.push(Edge(s, 0));
    while(!que.empty())
    {
        Edge tmp = que.top();
        int u = tmp.v;
        int cost = tmp.cost;
        que.pop();
        if(vis[u]) continue;
        vis[u] = true;
        term += dist[u];
        
        for(int i = eh[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].v;
            if(!vis[v] && dist[v] > edge[i].cost)
            {
                dist[v] = edge[i].cost;
                que.push(Edge(v, dist[v]));
            }
        }
    }
    return term;
}

void addedge(int u, int v, int cost)
{
    Edge e = Edge(v, cost, eh[u]);
    edge[tot] = e;
    eh[u] = tot++;
    return;
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    #endif
    while(scanf("%d", &nv) != EOF && nv)
    {
        memset(eh, -1, sizeof(eh));
        memset(vis, false, sizeof(vis));
        term = tot = 0;
        int u, v, cost, state;
        for(int i = 0; i < nv * (nv - 1) / 2; ++i)
        {
            scanf("%d%d%d%d", &u, &v, &cost, &state);
            if(state) cost = 0;
            addedge(u, v, cost);
            addedge(v, u, cost);
        }
        prim();
        printf("%d
", term);
    }
    return 0;
}

#include<cstdio>
#include<cstring>
#include<algorithm>
#define SIZE 102
#define MAXN 10000

using namespace std;

const int INF = 1<<30;
int nv, ne;
struct Edge{
    int u, v, cost;
}edge[MAXN];

int father[SIZE];

bool cmp(const struct Edge &a, const struct Edge &b)
{
    return a.cost < b.cost;
}

int find_father(int f)
{
    return father[f] = f == father[f] ? f : find_father(father[f]);
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    #endif
    while(scanf("%d", &nv) != EOF && nv)
    {
        int state;
        for(int i=1; i<=nv; ++i) father[i] = i;
        ne = nv*(nv-1)/2;
        for(int i=0; i<ne; ++i)
        {
            scanf("%d%d%d%d", &edge[i].u, &edge[i].v, &edge[i].cost, &state);
            if(state)
            {
                int u = find_father(edge[i].u);
                int v = find_father(edge[i].v);
                father[u] = v;
            }
        }
        int term = 0;
        sort(edge, edge+ne, cmp);
        for(int i=0; i<ne; ++i)
        {
            
            int u = find_father(edge[i].u);
            int v = find_father(edge[i].v);
            if(u != v) 
            {
                term += edge[i].cost;
                father[u] = v;
            }
        }
        printf("%d
", term);
    }
    return 0;
}