POJ 1815 Friendship ★(字典序最小点割集)
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题意】给出一个无向图，和图中的两个点s，t。求至少去掉几个点后才能使得s和t不连通，输出这样的点集并使其字典序最大。 不错的题，有助于更好的理解最小割和求解最小割的方法~ 【
思路】 问题模型很简单，就是无向图的点连通度，也就是最小点割集。麻烦之处在于需要使得点割集方案的字典序最大。这样的话通常的dfs划分点集的方法就行不通了。我们采取贪心的策略：枚举1~N的点，在残留网络中dfs检查其代表的边的两端点是否连通，如果不连通则该点可以为点割，那么我们就让他是点割，加入到答案中，然后删掉这条边更新最大流。重复这个过程直到扫描完所有点。 PS:很多人判断点是否可以是点割时是先删掉边然后判断流是否减小，我觉得这样是不是麻烦了？因为如果不减小的话（不是点割）还得把他还原，相当于重新构了两次图。
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#define MID(x,y) ((x+y)/2) #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; const int MAXV = 505; const int MAXE = 50005; const int oo = 0x3fffffff; /* Dinic-2.0-2013.07.21: adds template. double & int 转换方便多了，也不易出错 ~*/ template
struct Dinic{ struct node{ int u, v; T flow; int opp; int next; }arc[2*MAXE]; int vn, en, head[MAXV]; int cur[MAXV]; int q[MAXV]; int path[2*MAXE], top; int dep[MAXV]; void init(int n){ vn = n; en = 0; mem(head, -1); } void insert_flow(int u, int v, T flow){ arc[en].u = u; arc[en].v = v; arc[en].flow = flow; arc[en].next = head[u]; head[u] = en ++; arc[en].u = v; arc[en].v = u; arc[en].flow = 0; arc[en].next = head[v]; head[v] = en ++; } bool bfs(int s, int t){ mem(dep, -1); int lq = 0, rq = 1; dep[s] = 0; q[lq] = s; while(lq < rq){ int u = q[lq ++]; if (u == t){ return true; } for (int i = head[u]; i != -1; i = arc[i].next){ int v = arc[i].v; if (dep[v] == -1 && arc[i].flow > 0){ dep[v] = dep[u] + 1; q[rq ++] = v; } } } return false; } T solve(int s, int t){ T maxflow = 0; while(bfs(s, t)){ int i, j; for (i = 1; i <= vn; i ++) cur[i] = head[i]; for (i = s, top = 0;;){ if (i == t){ int mink; T minflow = 0x3fffffff; for (int k = 0; k < top; k ++) if (minflow > arc[path[k]].flow){ minflow = arc[path[k]].flow; mink = k; } for (int k = 0; k < top; k ++) arc[path[k]].flow -= minflow, arc[path[k]^1].flow += minflow; maxflow += minflow; top = mink; i = arc[path[top]].u; } for (j = cur[i]; j != -1; cur[i] = j = arc[j].next){ int v = arc[j].v; if (arc[j].flow && dep[v] == dep[i] + 1) break; } if (j != -1){ path[top ++] = j; i = arc[j].v; } else{ if (top == 0) break; dep[i] = -1; i = arc[path[-- top]].u; } } } return maxflow; } }; Dinic
dinic; bool map[MAXV][MAXV]; int n, s, t; bool vis[MAXV]; bool dfs(int u, int t){ vis[u] = 1; if (u == t) return true; for (int i = dinic.head[u]; i != -1; i = dinic.arc[i].next){ if (dinic.arc[i].flow <= 0) continue; int v = dinic.arc[i].v; if (!vis[v]){ if (dfs(v, t)){ return true; } } } return false; } bool del[MAXV]; void update_flow(){ dinic.init(2*n); for (int i = 1; i <= n; i ++){ if (del[i]) continue; if (i != s && i != t) dinic.insert_flow(i, n+i, 1); else{ dinic.insert_flow(i, n+i, oo); } for (int j = 1; j <= n; j ++){ if (del[j]) continue; if (i != j && map[i][j] == 1){ dinic.insert_flow(n+i, j, oo); } } } dinic.solve(s, t); } int main(){ //freopen("test.in", "r", stdin); //freopen("test.out", "w", stdout); while(scanf("%d %d %d", &n, &s, &t) != EOF){ dinic.init(2*n); bool if_answer = 1; for (int i = 1; i <= n; i ++){ if (i != s && i != t) dinic.insert_flow(i, n+i, 1); else{ dinic.insert_flow(i, n+i, oo); } for (int j = 1; j <= n; j ++){ scanf("%d", &map[i][j]); if (i != j && map[i][j] == 1){ dinic.insert_flow(n+i, j, oo); } if ((i == s && j == t && map[i][j] == 1)) if_answer = 0; } } if (!if_answer){ puts("NO ANSWER!"); continue; } int res = dinic.solve(s, t); printf("%d\n", res); vector
mincut; mem(del, false); if (res){ for (int i = 1; i <= n; i ++){ mem(vis, 0); if (!dfs(i, n+i)){ mincut.push_back(i); del[i] = true; update_flow(); } } for (int i = 0; i < (int)mincut.size(); i ++){ if (i == 0) printf("%d", mincut[i]); else printf(" %d", mincut[i]); } puts(""); } } return 0; }