POJ 3155 最大密度子图 二分+最小割
还是参考amber的论文 最小割那篇
这道题建图的话按照论文上建就可以。
但是对于本题来讲，最最蛋疼的地方绝对不是建图，而是精度。
比方说最后遍历残留网络的时候，因为是double类型么，我理所当然的用了eps去判断与0的关系，然后就杯具了。。。 交了十几次， 赫然发现尼玛直接写>0远比eps好用。
然后二分的部分也需要用到eps的时候，比如high与low的差大于eps，h(g)的值与eps比较，这样的eps从1e-2到1e-10均好使，什么，你想精度再高点？不好意思，再高就wa了
当然， high-low这一部分的eps可以用1.0/n/n来替代，这是一个定理
注意二分完了，一定要再用low建图求一遍最大流 否则还是wa 据说这是因为这个函数的图形非常奇特，经常会来个突然的变化
#include <iostream>#include <algorithm>#include <cstring>#include <string>#include <cstdio>#include <cmath>#include <queue>#include <map>#include <set>#define eps 1e-5#define MAXN 111#define MAXM 11111#define INF 1000000007using namespace std;struct node{ int ver; // vertex double cap; // capacity double flow; // current flow in this arc int next, rev;}edge[MAXM];int dist[MAXN], numbs[MAXN], src, des, n;int head[MAXN], e;void add(int x, int y, double c){ //e记录边的总数 edge[e].ver = y; edge[e].cap = c; edge[e].flow = 0; edge[e].rev = e + 1; //反向边在edge中的下标位置 edge[e].next = head[x]; //记录以x为起点的上一条边在edge中的下标位置 head[x] = e++; //以x为起点的边的位置 //反向边 edge[e].ver = x; edge[e].cap = 0; //反向边的初始容量为0 edge[e].flow = 0; edge[e].rev = e - 1; edge[e].next = head[y]; head[y] = e++;}void rev_BFS(){ int Q[MAXN], qhead = 0, qtail = 0; for(int i = 1; i <= n; ++i) { dist[i] = MAXN; numbs[i] = 0; } Q[qtail++] = des; dist[des] = 0; numbs[0] = 1; while(qhead != qtail) { int v = Q[qhead++]; for(int i = head[v]; i != -1; i = edge[i].next) { if(edge[edge[i].rev].cap == 0 || dist[edge[i].ver] < MAXN)continue; dist[edge[i].ver] = dist[v] + 1; ++numbs[dist[edge[i].ver]]; Q[qtail++] = edge[i].ver; } }}void init(){ e = 0; memset(head, -1, sizeof(head));}double maxflow(){ int u; double totalflow = 0; int Curhead[MAXN], revpath[MAXN]; for(int i = 1; i <= n; ++i)Curhead[i] = head[i]; u = src; while(dist[src] < n) { if(u == des) // find an augmenting path { double augflow = INF; for(int i = src; i != des; i = edge[Curhead[i]].ver) augflow = min(augflow, edge[Curhead[i]].cap); for(int i = src; i != des; i = edge[Curhead[i]].ver) { edge[Curhead[i]].cap -= augflow; edge[edge[Curhead[i]].rev].cap += augflow; edge[Curhead[i]].flow += augflow; edge[edge[Curhead[i]].rev].flow -= augflow; } totalflow += augflow; u = src; } int i; for(i = Curhead[u]; i != -1; i = edge[i].next) if(edge[i].cap > 0 && dist[u] == dist[edge[i].ver] + 1)break; if(i != -1) // find an admissible arc, then Advance { Curhead[u] = i; revpath[edge[i].ver] = edge[i].rev; u = edge[i].ver; } else // no admissible arc, then relabel this vertex { if(0 == (--numbs[dist[u]]))break; // GAP cut, Important! Curhead[u] = head[u]; int mindist = n; for(int j = head[u]; j != -1; j = edge[j].next) if(edge[j].cap > 0)mindist = min(mindist, dist[edge[j].ver]); dist[u] = mindist + 1; ++numbs[dist[u]]; if(u != src) u = edge[revpath[u]].ver; // Backtrack } } return totalflow;}int d[MAXN];int xx[MAXM], yy[MAXM];int nt, m;int vis[MAXN], ans;void dfs(int u){ vis[u] = 1; if(u <= nt) ans++; for(int i = head[u]; i != -1; i = edge[i].next) if(!vis[edge[i].ver] && edge[i].cap > 0) dfs(edge[i].ver);}void build(double mid){ for(int i = 1; i <= m; i++) { add(xx[i], yy[i], 1); add(yy[i], xx[i], 1); } for(int i = 1; i <= nt; i++) { add(src, i, m); add(i, des, m * 1.0 + 2 * mid - d[i] * 1.0); }}int main(){ while(scanf("%d%d", &nt, &m) != EOF) { if(m == 0) { printf("1\n1\n"); continue; } memset(d, 0, sizeof(d)); for(int i = 1; i <= m; i++) { scanf("%d%d", &xx[i], &yy[i]); d[xx[i]]++; d[yy[i]]++; } double low = 0, high = m; src = nt + 1; des = nt + 2; n = des; while(high - low > 1.0 / nt / nt) { init(); double mid = (low + high) / 2; build(mid); rev_BFS(); double h = (m * nt * 1.0 - maxflow()) / 2; if(h > eps) low = mid; else high = mid; } init(); build(low); rev_BFS(); maxflow(); memset(vis, 0, sizeof(vis)); ans = 0; dfs(src); printf("%d\n", ans); for(int i = 1; i <= nt; i++) if(vis[i]) printf("%d\n", i); } return 0;}