Exercise 6-14 The problem of inspectors UVa12118

1. Title Description: Click to open the link

2. Problem-solving ideas: Using DFS and Euler roads to solve problems. In fact, the need to find some path, so that the path and the required paths to form the Euler road (note is not Euler circuit, "Euler Road" refers to the path in addition to the beginning and end of the other points on the route is equal to the degree of the road). The first is to use DFS to determine connectivity, there is another reason to use DFS here is to count the number of points of odd degrees , because such a point will need to add a road so that its degree becomes even, in line with the requirements of Euler road. Since adding a path causes the two endpoints to become 2 degrees at the same time, and the number of start and end points is odd, the count is subtracted by 2, so the final path total is e+ (res-2)/2. where (res-2)/2 is the last number of paths added.

3. Code:

#define _crt_secure_no_warnings #include <iostream> #include <algorithm> #include <string> #include <sstream> #include <set> #include <vector> #include <stack> #include <map> #include < queue> #include <deque> #include <cstdlib> #include <cstdio> #include <cstring> #include < cmath> #include <ctime> #include <functional>using namespace std; #define N 1000+10int V, E, t;vector< Int>road[n];int vis[n];void init () {memset (Vis, 0, sizeof (VIS)), for (int i = 0; i < N; i++) road[i].clear ();} int dfs (int n) {if (Vis[n]) return 0;int cnt = 0;vis[n] = 1;cnt + = (road[n].size () & 1);//statistic degree is an odd point for (int i = 0; I &lt ; Road[n].size (); i++) cnt + = DFS (road[n][i]); return CNT;} int solve () {int res = 0;for (int i = 1; I <= v;i++) if (!vis[i] &&! Road[i].empty ())//Note Be sure to prevent repeated accumulation! Res + = max (Dfs (i), 2);//statistics of the number of points with odd degrees return t* (Max ((res-2)/2, 0) + e);//The final answer is E plus the number of newly added paths (res-2)/2}int main () {//freop En ("T.txt", "R", stdin); int rnd =0;while (Init (), CIN >> V >> E >> T, V | | E | | T) {int x, y;for (int i = 0; i < E; i++) {scanf ("%d%d", &x, &y); Road[x].push_back (y); Road[y].push_back (x);} printf ("Case%d:%d\n", ++rnd, Solve ());} return 0;}

Exercise 6-14 The problem of inspectors UVa12118