Poj 2942-Knights of the Round Table (dual connected map Tarjan + Binary Decision)

Poj 2942-Knights of the Round Table (dual connected map Tarjan + Binary Decision)

ACM

Question address:
Poj 2942-Knights of the Round Table

Question:
There are n knights, which provide the relationship of hatred between some of them. During the meeting, the knights will sit around a round table. A meeting can be held smoothly, meeting two conditions:

1. Any two servers that hate each other cannot be adjacent.
2. The number of participants is an odd number greater than 2.

If a server guard cannot participate in any meeting, he must be kicked out to give a hate relationship between the server guard and ask how many server guard should be kicked at least?

Analysis:
Take the server guard as a point, connect the server guard that can sit together, and then hold a meeting as a connected graph that only contains the odd circle. As long as you mark all the points that can hold the meeting, otherwise, t is needed.

The problem now is to find a connected graph that only contains odd circles. Note that it is a Round Table. Therefore, the number of nodes in a block is an odd number, not a odd circle. The odd circle is a ring with an odd number of nodes, there is a theorem: If a dual-connected component has an odd circle, the dual-connected component must not be a bipartite graph. They are necessary.

Therefore, first obtain a connected graph, then write a DFS binary to determine each connected graph, and then mark the qualified server guard.

Code:

/**  Author:      illuz <iilluzen[at]gmail.com>*  Blog:        http://blog.csdn.net/hcbbt*  File:        2942.cpp*  Create Date: 2014-08-15 16:30:10*  Descripton:  tarjan, bipartite*/#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <stack>using namespace std;#define repf(i,a,b) for(int i=(a);i<=(b);i++)typedef long long ll;const int N = 1010;struct Edge {int u;int v;};// cut points' bccno is no senseint dfn[N], iscut[N], bccno[N], tclock, bcccnt;int odd[N], color[N], A[N][N];int cas = 0, n, m, u, v;vector<int> G[N], bcc[N];stack<Edge> S;int tarjan(int u, int fa) {// current node and father nodeint lowu = dfn[u] = ++tclock;int child = 0;// number of child nodesint sz = G[u].size();repf (i, 0, sz - 1) {int v = G[u][i];Edge e = (Edge){u, v};if (!dfn[v]) {S.push(e);child++;int lowv = tarjan(v, u);lowu = min(lowu, lowv);if (lowv >= dfn[u]) {iscut[u] == true;bcc[++bcccnt].clear();// push the nodes to bccwhile (1) {Edge x = S.top();S.pop();if (bccno[x.u] != bcccnt) {bcc[bcccnt].push_back(x.u);bccno[x.u] = bcccnt;}if (bccno[x.v] != bcccnt) {bcc[bcccnt].push_back(x.v);bccno[x.v] = bcccnt;}if (x.u == u && x.v == v)break;}}} else if (dfn[v] < dfn[u] && v != fa) {// undirected so v != faS.push(e);lowu = min(lowu, dfn[v]);}}if (fa < 0 && child == 1)// if u is rootiscut[u] = 0;return lowu;}void find_bcc(int n) {memset(dfn, 0, sizeof(dfn));memset(iscut, 0, sizeof(iscut));memset(bccno, 0, sizeof(bccno));tclock = bcccnt = 0;repf (i, 0, n - 1)if (!dfn[i])tarjan(i, -1);}bool bipartite(int u, int b) {int sz = G[u].size();repf (i, 0, sz - 1) {int v = G[u][i];if (bccno[v] != b)continue;if (color[v] == color[u])return false;if (!color[v]) {color[v] = 3 - color[u];if (!bipartite(v, b))return false;}}return true;}void read() {repf (i, 0, n - 1)G[i].clear();memset(A, 0, sizeof(A));repf (i, 0, m - 1) {scanf("%d%d", &u, &v);u--;v--;A[u][v] = A[v][u] = 1;}repf (u, 0, n - 1)repf (v, u + 1, n - 1) {if (!A[u][v]) {G[u].push_back(v);G[v].push_back(u);}}}void solve() {find_bcc(n);memset(odd, 0, sizeof(odd));repf (i, 1, bcccnt) {memset(color, 0, sizeof(color));int sz = bcc[i].size();repf (j, 0, sz - 1) {bccno[bcc[i][j]] = i;}int u = bcc[i][0];color[u] = 1;if (!bipartite(u, i))repf (j, 0, sz - 1) {odd[bcc[i][j]] = 1;}}int ans = n;repf (i, 0, n - 1)if (odd[i])ans--;printf("%d\n", ans);}int main() {while (~scanf("%d%d", &n, &m) && n) {read();solve();}return 0;}