POJ3164-Command Network

最新推荐文章于 2018-10-30 11:40:59 发布

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最新推荐文章于 2018-10-30 11:40:59 发布

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分类专栏： POJ ----生成树

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探讨在战争环境下如何快速建立一个有效的单向通信网络，确保指挥部能够迅速传递指令到每一个节点。文章提出了一种算法，旨在寻找最短路径的有向树形结构，使总体线路长度最短。

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Command Network

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 17829		Accepted: 5111

Description

After a long lasting war on words, a war on arms finally breaks out between littleken’s and KnuthOcean’s kingdoms. A sudden and violent assault by KnuthOcean’s force has rendered a total failure of littleken’s command network. A provisional network must be built immediately. littleken orders snoopy to take charge of the project.

With the situation studied to every detail, snoopy believes that the most urgent point is to enable littenken’s commands to reach every disconnected node in the destroyed network and decides on a plan to build a unidirectional communication network. The nodes are distributed on a plane. If littleken’s commands are to be able to be delivered directly from a node A to another node B, a wire will have to be built along the straight line segment connecting the two nodes. Since it’s in wartime, not between all pairs of nodes can wires be built. snoopy wants the plan to require the shortest total length of wires so that the construction can be done very soon.

Input

The input contains several test cases. Each test case starts with a line containing two integer N (N ≤ 100), the number of nodes in the destroyed network, and M (M ≤ 10⁴), the number of pairs of nodes between which a wire can be built. The next N lines each contain an ordered pair x_i and y_i, giving the Cartesian coordinates of the nodes. Then follow M lines each containing two integers iand j between 1 and N (inclusive) meaning a wire can be built between node i and node j for unidirectional command delivery from the former to the latter. littleken’s headquarter is always located at node 1. Process to end of file.

Output

For each test case, output exactly one line containing the shortest total length of wires to two digits past the decimal point. In the cases that such a network does not exist, just output ‘poor snoopy’.

Sample Input

Sample Output

31.19
poor snoopy

Source

POJ Monthly--2006.12.31, galaxy

题意：有N个点，M条有向边。以1为根构造出一棵最小生成树，问这棵最小生成树能否被构造出来，如果可以，总权值是多少
解题思路：用的是double型的，输出时是%.2lf，结果是WA ，换成了%.2f就A了

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;
const int MAXN = 1010;
const int MAXM = 40010;

struct Edge
{
    int u,v;
    double cost;
}edge[MAXM];

struct Point
{
    double x,y;
}p[105];

int pre[MAXN],id[MAXN],visit[MAXN];
double in[MAXN];

double zhuliu(int root,int n,int m,Edge edge[])
{
    double res = 0;
    int u,v;
    while(1)
    {
        for(int i = 0; i < n; i++) in[i] = 1.0*INF;
        for(int i = 0; i < m; i++)
        {
            if(edge[i].u != edge[i].v && edge[i].cost < in[edge[i].v])
            {
                pre[edge[i].v] = edge[i].u;
                in[edge[i].v] = edge[i].cost;
            }
        }
        for(int i = 0; i < n; i++)
            if(i != root && in[i] == 1.0*INF) return -1; //不存在最小树形图
        int tn = 0;
        memset(id,-1,sizeof(id));
        memset(visit,-1,sizeof(visit));
        in[root] = 0;
        for(int i = 0; i < n; i++)
        {
            res += in[i];
            v = i;
            while( visit[v] != i && id[v] == -1 && v != root)
            {
                visit[v] = i;
                v = pre[v];
            }
            if( v != root && id[v] == -1 )
            {
                for(u = pre[v]; u != v ; u = pre[u])
                    id[u] = tn;
                id[v] = tn++;
            }
        }
        if(tn == 0) break;//没有有向环
        for(int i = 0; i < n; i++)
            if(id[i]==-1) id[i]=tn++;
        for(int i = 0; i < m;i++)
        {
            v = edge[i].v;
            edge[i].u = id[edge[i].u];
            edge[i].v = id[edge[i].v];
            if(edge[i].u != edge[i].v) edge[i].cost -= in[v];
            //else swap(edge[i],edge[--m]);
        }
        n = tn;
        root = id[root];
    }
    return res;
}

int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=0;i<n;i++)
            scanf("%lf %lf",&p[i].x,&p[i].y);
        int u,v;
        int L = 0;
        while(m--)
        {
            scanf("%d%d",&u,&v);
            u--,v--;
            double dis=1.0*sqrt((p[u].x-p[v].x)*(p[u].x-p[v].x)+(p[u].y-p[v].y)*(p[u].y-p[v].y));
            edge[L].u = u;
            edge[L].v = v;
            edge[L++].cost =dis;
        }
        double ans=zhuliu(0,n,L,edge);
        if(ans==-1) printf("poor snoopy\n");
        else printf("%.2lf\n",ans);
    }
    return 0;
}