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| Time Limit: 1000MS | Memory Limit: 131072K | |
| Total Submissions: 14782 | Accepted: 4249 |
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
4 6 0 6 4 6 0 0 7 20 1 2 1 3 2 3 3 4 3 1 3 2 4 3 0 0 1 0 0 1 1 2 1 3 4 1 2 3
Sample Output
31.19poor snoopy
题意：给你N个点的坐标和M条有向边，问你以点1为根的最小树形图的边权之和。
入门题：因为犯二把缩点那部分放到记录环的for循环里面，找bug找了好久。
#include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <algorithm> #define MAXN 110 #define MAXM 20000+10 #define INF 0x3f3f3f3f using namespace std; struct Edge { int from, to; double cost; }; Edge edge[MAXM]; int pre[MAXN]; int vis[MAXN]; int id[MAXN]; double in[MAXN]; int N, M; struct Node { double x, y; }; Node num[MAXN]; double dis(Node a, Node b) { return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)); } void getMap() { for(int i = 0; i < N; i++) scanf("%lf%lf", &num[i].x, &num[i].y); int a, b; for(int i = 0; i < M; i++) { scanf("%d%d", &a, &b); a--, b--; edge[i].from = a; edge[i].to = b; edge[i].cost = a==b ? INF : dis(num[a], num[b]);//去除自环 自己到自己无穷大 } } 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] = INF; for(int i = 0; i < m; i++) { Edge E = edge[i]; if(E.from != E.to && E.cost < in[E.to]) { pre[E.to] = E.from; in[E.to] = E.cost; } } for(int i = 0; i < n; i++) if(i != root && in[i] == INF) return -1; int tn = 0; memset(id, -1, sizeof(id)); memset(vis, -1, sizeof(vis)); in[root] = 0; for(int i = 0; i < n; i++) { res += in[i]; v = i; while(vis[v] != i && id[v] == -1 && v != root) { vis[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; for(int i = 0; i < n; i++) if(id[i] == -1) id[i] = tn++; for(int i = 0; i < m; ) { v = edge[i].to; edge[i].from = id[edge[i].from]; edge[i].to = id[edge[i].to]; if(edge[i].from != edge[i].to) edge[i++].cost -= in[v]; else swap(edge[i], edge[--m]); } n = tn; root = id[root]; } return res; } int main() { while(scanf("%d%d", &N, &M) != EOF) { getMap(); double ans = zhuliu(0, N, M, edge); if(ans == -1) printf("poor snoopy\n"); else printf("%.2lf\n", ans); } return 0; }
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poj 3164 Command Network 【最小树形图】【朱刘算法 入门】
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原文地址：http://blog.csdn.net/chenzhenyu123456/article/details/48000025
