POJ 3164 Command Network

POJ 3164 Command Network

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 integerN (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 nextN lines each contain an ordered pair x_i and y_i, giving the Cartesian coordinates of the nodes. Then followM lines each containing two integers i and j between 1 andN (inclusive) meaning a wire can be built between node i and nodej 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

4 60 64 60 07 201 21 32 33 43 13 24 30 01 00 11 21 34 12 3

Sample Output

31.19poor snoopy

Source

POJ Monthly--2006.12.31, galaxy

Solution

朱刘算法，与HDU 2121 Ice_cream’s world II类似，只是这道题的根节点是０

Code

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <stack>#include <map>#include <vector>#include <queue>#define L 200#define LL long long#define inf 2000000000using namespace std;struct point {  double x, y;} p[L];struct node {  int from, to;  double w;} e[L * L];int n, m, pre[L], vis[L], id[L];double in[L];inline double len(int a, int b) {  return sqrt(a * a + b * b);}inline double MST(int root, int N, int E) {  double ret = 0;  while (true) {    for (int i = 0; i < N; ++i) in[i] = inf;    for (int i = 0; i < E; ++i) {      int u = e[i].from, v = e[i].to;      double w = e[i].w;      if (w < in[v] && v != u) in[v] = w, pre[v] = u;    }    for (int i = 0; i < N; ++i) {      if (i == root) continue;      if (in[i] == inf) return -1;    }    in[root] = 0;    memset(vis, -1, sizeof(vis));    memset(id, -1, sizeof(id));    int node = 0;    for (int i = 0; i < N; ++i) {      ret += in[i];      int v = i;      while (vis[v] != i && v != root && id[v] == -1) vis[v] = i, v = pre[v];      if (v != root && id[v] == -1) {for (int u = pre[v]; u != v; u = pre[u]) id[u] = node;id[v] = node++;      }    }    if (node == 0) break;    for (int i = 0; i < N; ++i)      if (id[i] == -1) id[i] = node++;    for (int i = 0; i < E; ++i) {      int u = e[i].from, v = e[i].to;      e[i].from = id[u], e[i].to = id[v];      if (e[i].from != e[i].to) e[i].w -= in[v];     }    N = node, root = id[root];  }  return ret;}int main() {  freopen("POJ3164.in", "r", stdin);  freopen("POJ3164.out", "w", stdout);  while (scanf("%d %d", &n, &m) != EOF) {    for (int i = 0; i < n; ++i) scanf("%lf %lf", &p[i].x, &p[i].y);    for (int i = 0; i < m; ++i) {      scanf("%d %d", &e[i].from, &e[i].to);      e[i].from--, e[i].to--;      if (e[i].from != e[i].to) e[i].w = len(p[e[i].from].x - p[e[i].to].x, p[e[i].from].y - p[e[i].to].y);      else e[i].w = inf;    }    double ans = MST(0, n, m);    if (ans == -1) printf("poor snoopy\n");    else printf("%.2f\n", ans);  }  return 0;}

Summary

一开始本来都是对的，只是

printf("%.2f\n", ans);

打成了

printf("%.2lf\n", ans);

就WA了……