http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2676
大致题意：给出一个带权无向图，每条边有一个边权wi,求将S和T分开的一个割边集C，使得该割边集的平均边权最小，即最小化∑wi / |C| 。
思路：amber论文中详细讲解了如何转化成函数及建图，值得注意的是当边被重新赋权后,对于wi < 0 的边权，该边必然在最小割中，不必再建边，直接加入最大流中即可，因为求最小割时边权都为正值。
最后输出的是所选割边的序号。求割边无非是从源点dfs,每次走残量网络中流量大于0的边并标记端点，最后判断边的两个端点一个标记一个未标记，那么该边便是割边。
这题我TLE了13次，最后是因为Dinic的原因，可能之前的那个模板耗时太长了。改成了学长的Dinic,瞬间就A了。
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#define LL long long
#define _LL __int64
#define eps 1e-7
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 210;
const int maxm = 6000;
int s,t;
struct node
{
int u,v;
double w;
int next;
int re;
}p[maxm],edge[maxm];
int n,m;
int cnt,head[maxn];
int dist[maxn],vis[maxn];
void init()
{
cnt = 0;
memset(head,-1,sizeof(head));
}
void add(int u, int v, double w)
{
edge[cnt] = (struct node){u,v,w,head[u],cnt+1};
head[u] = cnt++;
edge[cnt] = (struct node){v,u,0,head[v],cnt-1};
head[v] = cnt++;
}
bool bfs()
{
queue <int> que;
memset(dist, 0, sizeof(dist));
memset(vis, 0, sizeof(vis));
while(!que.empty()) que.pop();
vis[s] = 1;
que.push(s);
while(!que.empty())
{
int u = que.front();
que.pop();
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].v;
if(edge[i].w && !vis[v])
{
que.push(v);
vis[v] = 1;
dist[v] = dist[u]+1;
}
}
}
if(dist[t] == 0)
return false;
return true;
}
double dfs(int u, double delta)
{
if(u == t) return delta;
double ret = 0,tmp;
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].v;
if(edge[i].w && dist[edge[i].v] == dist[u]+1 && (tmp = dfs(v,min(delta,edge[i].w))))
{
edge[i].w -= tmp;
edge[edge[i].re].w += tmp;
return tmp;
}
}
if(!ret) dist[u] = -1;
return ret;
}
double Dinic()
{
double ret = 0,res;
while(bfs())
{
while(res = dfs(s,INF))
ret += res;
}
return ret;
}
bool ok(double mid)
{
init();
double flow = 0;
for(int i = 1; i <= m; i++)
{
if(p[i].w > mid)
{
add(p[i].u,p[i].v,p[i].w-mid);
add(p[i].v,p[i].u,p[i].w-mid);
}
else
flow += p[i].w - mid;
}
flow += Dinic();
if(flow > eps)
return true;
else return false;
}
void dfs_cut(int u)
{
vis[u] = 1;
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].v;
if(!vis[v] && edge[i].w > 0)
{
dfs_cut(v);
}
}
}
int main()
{
int item = 0;
while(~scanf("%d %d",&n,&m))
{
s = 1;
t = n;
item += 1;
if(item >= 2)
printf("\n");
double low = INF,high = 0,mid;
for(int i = 1; i <= m; i++)
{
scanf("%d %d %lf",&p[i].u, &p[i].v, &p[i].w);
low = min(low,p[i].w);
high = max(high,p[i].w);
}
while( fabs(high - low) > eps)
{
mid = (high + low)/2.0;
if( ok(mid) )
low = mid;
else high = mid;
}
memset(vis,0,sizeof(vis));
dfs_cut(1);
int count = 0;
int ans[maxm];
for(int i = 1; i <= m; i++)
{
if(vis[p[i].u] + vis[p[i].v] == 1 || p[i].w <= mid)
ans[++count] = i;
}
printf("%d\n",count);
for(int i = 1; i <= count-1; i++)
printf("%d ",ans[i]);
printf("%d\n",ans[count]);
}
return 0;
}