題意：找出圖中不可能在奇圈中的點. [分析]注意到，在不同點雙連通分量中的兩個點，顯然是不會存在圈的.那么這樣，問題就劃歸為在點雙連通分量中去找奇圈。 [重要性質]在一個點雙連通分量中,只要有任意一個奇圈，那么所有的點都可以在一個奇圈內(證明看《算法競賽入門經典 訓練指南》). [重要定理]一個圖含奇圈當且僅當圖不是二分圖. [解題思路]先求出圖的點雙連通分量(塊),然后對每一個塊染色判斷二分圖,統(tǒng)計出不可能在奇圈中的點的個數 [注意]染色判定二分圖的算法要寫對 ? #include #include #include #include #include #include #include #include #include #include #include #include #define MID(x,y) ((x+y)>>1) #define mem(a,b) memset(a,b,sizeof(a)) using namespace std;const int N = 1002; const int E = 2000002; struct node{int u, v;int next; }arc[E]; int cnt, head[N]; void init(){mem(head, -1);cnt = 0; } void add(int u, int v){arc[cnt].u = u;arc[cnt].v = v;arc[cnt].next = head[u];head[u] = cnt ++;arc[cnt].u = v;arc[cnt].v = u;arc[cnt].next = head[v];head[v] = cnt ++; }/* 求點雙連通分量 */ int dfn[N], low[N]; set bcc[N]; int id, bcc_num; stack st; void addbcc(int u, int v){bcc[bcc_num].insert(u);bcc[bcc_num].insert(v); } void dfs(int u, int father){dfn[u] = low[u] = ++id;for (int i = head[u]; i != -1; i = arc[i].next){int v = arc[i].v;if (v == father) continue;if (dfn[v] < dfn[u]){st.push(arc[i]);if (!dfn[v]){dfs(v, u);low[u] = min(low[u], low[v]);if (dfn[u] <= low[v]){++ bcc_num;while(!st.empty()){int a = st.top().u;int b = st.top().v;st.pop();addbcc(a, b);if ((a == u && b == v) || (b == u && a == v) )break;}}}else{low[u] = min(low[u], dfn[v]);}}} } void bcc_tarjan(int n){id = bcc_num = 0;mem(dfn, 0);mem(low, 0);while(!st.empty())st.pop();for (int i = 0; i <= n; i ++)bcc[i].clear();for (int i = 1; i <= n; i ++)dfs(i, 0); } /* 求點雙連通分量 *//* 染色判定二分圖 */ int col[N]; bool not_bigragh[N]; //標記某個點雙連通分量是不是二分圖 void dfs_color(int bcc_id, int u, int color){col[u] = color;for (int i = head[u]; i != -1; i = arc[i].next){if (not_bigragh[bcc_id])return;int v = arc[i].v;if (bcc[bcc_id].find(v) == bcc[bcc_id].end())continue;if (col[v] == col[u]){not_bigragh[bcc_id] = 1;return;}else if (col[v] == -1){dfs_color(bcc_id, v, (color+1)&1);}} } bool fill(int bcc_id){ //對某個點雙連通分量染色判斷二分圖set ::iterator it;for (it = bcc[bcc_id].begin(); it != bcc[bcc_id].end(); it ++){not_bigragh[bcc_id] = 0;int u = *it;mem(col, -1);dfs_color(bcc_id, u, 0);if (not_bigragh[bcc_id])return false;}return true; } /* 染色判定二分圖 */int res; bool can[N]; //存某個點能否在一個奇圈中 int solve(int n){res = 0;mem(can, 0);for (int i = 1; i <= bcc_num; i ++){if (!fill(i)){set ::iterator it;for (it = bcc[i].begin(); it != bcc[i].end(); it ++){can[*it] = 1;}}}for (int i = 1; i <= n; i ++){if (!can[i])res ++;}return res; } bool mat[N][N]; //表示i憎恨j的關系矩陣 int main(){int n, m;while(scanf("%d %d", &n, &m) != EOF){if (n == 0 && m == 0)break;init();mem(mat, 0);for (int i = 0; i < m; i ++){int a, b;scanf("%d %d", &a, &b);mat[a][b] = 1;mat[b][a] = 1;}for (int i = 1; i <= n; i ++){for (int j = i+1; j <= n; j ++){if (!mat[i][j]){add(i, j);}}}bcc_tarjan(n);printf("%d\n", solve(n));}return 0; } ?

轉載于:https://www.cnblogs.com/AbandonZHANG/archive/2013/06/05/4114246.html

總結

以上是生活随笔為你收集整理的POJ 2942 Knights of the Round Table ★(点双连通分量+二分图判定)的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。