POJ 3155 Hard Life（最大密度子图）

详细解法参考胡伯涛的《最小割模型在信息学竞赛中的应用 》论文，下面做了一些注释，仅供参考。。。

代码如下：

#include<iostream>#include<cstring>#include<cstdio>#include<queue>const int maxn = 1500;const int maxm = 10000;const double eps = 1e-6;const int INF = 0x3f3f3f3f;using namespace std;inline int read(){int x=0,t=1,c;while(!isdigit(c=getchar()))if(c=='-')t=-1;while(isdigit(c))x=x*10+c-'0',c=getchar();return x*t;}int n, m;int head[maxn],cur[maxn],nx[maxm<<1],to[maxm<<1],ppp=0,thead[maxn];double flow[maxm<<1];struct Dinic {int dis[maxn];int s, t;double ans;void init() {memset(head, -1, sizeof(head));ppp = 0;}void AddEdge(int u, int v, double c){to[ppp]=v;flow[ppp]=c;nx[ppp]=head[u];head[u]=ppp++;swap(u,v);to[ppp]=v;flow[ppp]=0;nx[ppp]=head[u];head[u]=ppp++;}bool BFS(){memset(dis, -1, sizeof(dis));dis[s] = 1; queue<int> Q;Q.push(s);while(!Q.empty()){int x = Q.front();Q.pop();for(int i = head[x]; ~i; i = nx[i]){if(flow[i] > 0 && dis[to[i]] == -1){dis[to[i]] = dis[x] + 1;Q.push(to[i]);}}}return dis[t] != -1;}double DFS(int x, double maxflow) {if(x == t || !maxflow){ans += maxflow;return maxflow;}double ret = 0;double f;for(int &i = cur[x]; ~i; i = nx[i]) {if(dis[to[i]] == dis[x] + 1 && (f = DFS(to[i], min(maxflow, flow[i])))) {ret += f;flow[i] -= f;flow[i^1] += f;maxflow -= f;if(!maxflow)break;}}return ret;}double solve(int source, int tank) {s = source;t = tank;ans = 0;while(BFS()) {memcpy(cur, head, sizeof(cur));DFS(s, INF);}return ans;}}dinic;int d[maxn], u[maxn], v[maxn], cnt; bool vst[maxn];void build(double g) {dinic.init();int s = 0, t = n + 1;for(int i = 1; i <= n; i++) {dinic.AddEdge(s, i, m);dinic.AddEdge(i, t, m + 2 * g - d[i]);}for(int i = 1; i <= m; i++) {dinic.AddEdge(u[i], v[i], 1.0);dinic.AddEdge(v[i], u[i], 1.0);}}void find_dfs(int uu) {cnt++;vst[uu] = 1;for(int i = head[uu]; ~i; i = nx[i]) {int vv = to[i];if(flow[i] > eps && !vst[vv]) {find_dfs(vv);}}}int main() {#ifndef ONLINE_JUDGEfreopen("poj_in.txt", "r", stdin);#endifn = read(), m = read();if(m == 0) {printf("1\n1\n");return 0;}for(int i = 1; i <= m; i++) {u[i] = read(), v[i] = read();d[u[i]]++, d[v[i]]++;}double front = 0, back = m;double e = 1.0 / n / n;     //引理4.1 无向图G中, while(back - front >= e) {  //任意两个具有不同密度的子图G1,G2double g = (front + back) / 2.0;//它们的密度差不小于1/n/n build(g);double tmp = dinic.solve(0, n + 1);if((n * m - tmp) / 2.0 > eps) //即使两边乘以-1，依然要趋于0 front = g;//大于0说明g还不够大,使得tmp太小 elseback = g;}build(front);     //得到比例后，重新建图 dinic.solve(0, n + 1);//没有流量的是红色的边 find_dfs(0);//故判断某条边是否有流量 printf("%d\n", cnt - 1);//即可知道某点是否在集合中 for(int i = 1; i <= n; i++) {if(vst[i])printf("%d\n", i);}return 0;}