TJU 2248. minimum tree structure of Channel Design, tju Tree Structure

TJU 2248. minimum tree structure of Channel Design, tju Tree Structure

Minimum tree structure, test template ....

2248. Channel DesignTime Limit: 1.0 Seconds Memory Limit: 65536 K
Total Runs: 2199 Accepted Runs: 740

We need irrigate our farms, but there is only one source of water nearby. So we need build some water channels with minimum cost.

In Figure (A),V_{1 indicates the source of water. OtherN-1 nodes in the Figure indicate the farms we need to irrigate. an edge represents you can build a channel between the two nodes, to irrigate the target. the integers indicate the cost of a channel between two nodes.}

Figure (B) Represents a design of channels with minimum cost.

Input

There are multiple cases, the first line of each case contains two integersNAndM(2 ≤N≤ 100; 1 ≤M≤ 10000 ),NShows the number of nodes. The followingMLines, each line contains three integersI j c_Ij, Means we can build a channel from nodeV_ITo nodeV_J, Which costC_Ij. (1 ≤I,J≤N;I=J; 1 ≤C_Ij≤ 100)

The source of water is alwaysV_{1.
The input is terminatedN=M= 0.}

Output

For each case, output a single line contains an integer represents the minimum cost.

If no design can irrigate all the farms, output "impossible" instead.

Sample Input

5 81 2 31 3 52 4 23 1 53 2 53 4 43 5 75 4 33 31 2 31 3 53 2 10 0

Sample Output

176

Problem setter:Hill

Source:  TJU Contest August 2006 Submit List Runs Forum Statistics

/*************************************** * ******** Author: CKbossCreated Time:, Saturday, July 04, 2015 File Name: TJU2248.cpp *************************************** * ********/# include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <string> # include <cmath> # include <cstdlib> # include <vector> # include <queue> # include <set> # include <map> using namespace std; const int INF = 0x3f3 F3f3f; const int maxn = 110; int n, m; struct Edge {int u, v, cost;}; Edge edge [maxn * maxn]; int pre [maxn], id [maxn], vis [maxn], in [maxn]; int zhuliu (int root, int n, int m, Edge edge []) {int res = 0, u, v; while (true) {for (int I = 0; I <n; I ++) in [I] = 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] = 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 (int 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]. 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]; elseswap (edge [I], edge [-- m]);} n = tn; root = id [root];} return res ;} int main () {// freopen ("in.txt", "r", stdin); // freopen ("out.txt", "w", stdout ); while (scanf ("% d", & n, & m )! = EOF) {if (n = 0 & m = 0) break; for (int I = 0; I <m; I ++) {int u, v, w; scanf ("% d", & u, & v, & w); u --; v --; edge [I]. u = u; edge [I]. v = v; edge [I]. cost = w;} int ans = zhuliu (0, n, m, edge); if (ans =-1) puts ("impossible "); else printf ("% d \ n", ans);} return 0 ;}

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