POJ 3164 Command Network（朱刘算法）

Command Network

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 21481		Accepted: 6117

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 xi and yi, giving the Cartesian coordinates of the nodes. Then follow M lines each containing two integers i and 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

unidirectional - > 单向 undirectional -> 双向

此题是要求有向图的最小生成树 - > 用朱刘算法

细节：double ，点的坐标也要用double 不然 Wa C++ AC G++ WA

#include <iostream>
#include <algorithm>
#include <stack>
#include <queue>
#include <map>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <iomanip>
#include <cmath>
#define inf 0x3f3f3f3f
#define Max 101
struct Node{double x, y;
}node[Max];
struct Edge{int u, v;double w;
}edge[Max*Max];
int vis[Max];
int id[Max];//结点所属环编号
double in[Max];
int pre[Max];//in[]为最小入边权,pre[]为其对应的起点
double zhuLiu(int root, int n, int m){//root结点、点数、边数double res = 0;//最小树形图总权值while(true){for(int i = 0; i < n; i++)//初始化为无穷大in[i] = inf;//寻找每个点的最小入边for(int i = 0; i < m; i++){//遍历每条边int x = edge[i].u;int y = edge[i].v;if(edge[i].w < in[y] && x != y){//更新最小入边pre[y] = x;//记录前驱in[y] = edge[i].w;//更新}}//判断是否存在最小树形图for(int i = 0; i < n; i++){if(i == root)continue;if(in[i] == inf)//除根节点外的点存在孤立点return -1;}//寻找所有的环int cnt = 0;//记录环数in[root] = 0;memset(id, -1, sizeof(id));memset(vis, -1, sizeof(vis));for(int i = 0; i < n; i++){//标记每个环res += in[i];//记录权值int y = i;while(vis[y] != i && id[y] == -1 && y != root){//寻找图中有向环//三种情况会终止：找到出现同样标记的点、结点已属其他环、遍历到根vis[y] = i;//标记y = pre[y];//向上找}if(y != root && id[y] == -1){//没有遍历到根或没有找到结点属于其他环,说明找到有向环for(int x = pre[y]; x != y; x = pre[x])//标记结点x为第几个环id[x] = cnt;//记录结点所属环号id[y] = cnt++;//记录结点所属环号并累加}}if(cnt == 0)//无环break;for(int i = 0; i < n; i++)//可能存在独立点if(id[i] == -1)//环数累加id[i] = cnt++;//建立新图,缩点重新标记for(int i = 0; i < m; i++){int x = edge[i].u;int y = edge[i].v;edge[i].u = id[x];edge[i].v = id[y];if(id[x] != id[y])//两点不在同一环内,更新边权值edge[i].w -= in[y];//x到y的距离为边权-in[y]}n = cnt;//以环数为下次操作的点数,继续上述操作,直到无环root = id[root];}return res;
}
// int main(){
//     int n,m;//n个点m条有向边
//     scanf("%d%d",&n,&m);
//     for(int i=0;i<m;i++){//建图
//         scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].w);
//         if(edge[i].x==edge[i].y)//除去自环,即点到自身距离为INF
//             edge[i].w=INF;
//     }
//     int res=zhuLiu(0,n,m);
//     if(res==-1)
//         printf("No\n");
//     else
//         printf("%d\n",res);
//     return 0;
// }
double dis(Node a,Node b) {return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));
}
int main() {int n, m, u, v;while(scanf("%d %d",&n, &m) != EOF) {for(int i = 0; i < n; i++) {scanf("%lf %lf",&node[i].x, &node[i].y);}for(int i = 0; i < m; i++) {scanf("%d %d", &u, &v);u--;v--;edge[i].u = u; edge[i].v = v;edge[i].w = dis(node[u], node[v]);if(edge[i].u == edge[i].v) {edge[i].w = inf*1.0;}}double ans = zhuLiu(0, n, m);if(ans == -1) printf("poor snoopy\n");else printf("%.2lf\n",ans);}return 0;
}

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