题意：给定无向图，每条边有权值，求该图的一个割集，是的该割集的平均边权最小
Amber的《最小割模型在信息学竞赛中的应用》中讲的很清楚了。
二分答案k，对每条边进行重新赋值为原边权-k，求最大流，
可看这里：http://hi.baidu.com/buaa_babt/item/a08fbb45599dc722fb89602a
二分枚举当前的平均边长l，对于边权<=l的直接加入当前最优割集，边权>l的将容量设为边权-l，加入到网络中，求出最小割的和sum，sum加上刚刚那些小于l的边(也是边权-l)，如果大于0就意味着l不可作为最小平均边权，小于0就可以，当等于0时直接输出即可。一定要注意精度
注意是无向图，还有就是求st割，求解答案。
#define inf 0x3f3f3f3f #define maxm 100000 #define maxn 10000 #define eps 1e-9 struct node { int u ; int v,next,id; double w; void init(int u1,int v1,double w1) { u = u1; v = v1; w = w1; } }; node e1[maxn]; int head[maxn]; int cnt ; int cnt1; node edge[maxn]; int sgn(double x) { //return -1 0 1 return x < -eps ? -1 : x > eps; } void init() { memset(head,-1,sizeof(head)); cnt = 0 ; } void add(int u,int v,int id,double w) { edge[cnt].u = u; edge[cnt].v = v; edge[cnt].w = w; edge[cnt].id = id; edge[cnt].next = head[u]; head[u] = cnt ++ ; edge[cnt].u = v; edge[cnt].v = u; edge[cnt].w = 0; edge[cnt].id = id; edge[cnt].next = head[v]; head[v] = cnt ++ ; } int dis[maxn]; int pre[maxn],cur[maxn],gap[maxn]; int s,t,nn; int sap() { double flow,aug; int u; int flag; int i; flow=0; aug=inf; for(i=0; i<nn; i++) { cur[i]=head[i]; gap[i]=0; dis[i]=0; } gap[s]=nn; u=s; pre[s]=s; while(dis[s]<nn) { flag=0; for(int &j=cur[u]; j!=-1; j=edge[j].next) { //cur[u]=j; int v=edge[j].v; if(edge[j].w>0&&dis[u]==dis[v]+1) { flag=1; if(edge[j].w<aug)aug=edge[j].w; pre[v]=u; u=v; if (u==t) { flow+=aug; while(u!=s) { u=pre[u]; edge[cur[u]].w-=aug; edge[cur[u]^1].w+=aug; //why?解释偶数异或1为偶数+1，奇数异或1为奇数-1， //显然我们存的边是从0开始存的， //所以偶数,偶数+1是残量网格中的两条边(无向边) } aug=inf; } break; } } if (flag) continue; int mindis=nn; for(int j=head[u]; j!=-1; j=edge[j].next) { int v=edge[j].v; if (edge[j].w>0&&dis[v]<mindis) { mindis=dis[v]; cur[u]=j; } } if (--gap[dis[u]]==0)//间隙优化 { break; } dis[u]=mindis+1; gap[dis[u]]++; u=pre[u]; } return flow; } bool xxx[maxn]; int n,m; double judge(double lda) { memset(xxx,0,sizeof(xxx)); init(); double res = 0.0; for(int i = 1; i <= m ; i ++) { double tmp ; tmp = e1[i].w - lda; if( tmp >=0 ) add(e1[i].u,e1[i].v,i,tmp) , add(e1[i].v,e1[i].u,i,tmp); // 注意 else { res += tmp; xxx[i] = 1 ; } //if(sgn(tmp) < 0 ) // res += tmp ; //else add(e1[i].u,e1[i].v,i,tmp); } s = 1; t = n ; nn = t + 2; res += sap(); return res; } queue<int> q; int col[maxn]; bool vis[maxn]; void bfs(int st,int color) { memset(vis,0,sizeof(vis)); memset(col,0,sizeof(col)); while(!q.empty()) q.pop(); q.push(st); vis[st] = 1; col[st] = color; while(!q.empty()) { int u = q.front(); q.pop(); for(int i = head[u] ; i != -1; i = edge[i].next) { int v = edge[i].v; if(sgn(edge[i].w) > 0 && vis[v] == 0 ) { vis[v] = 1; col[v] = color; q.push(v); } } } return ; } vector<int> ans ; void getst(double lda) { bfs(1,1); ans.clear(); for(int i = 1 ; i <= m ; i ++ ) // 注意 { if(!xxx[i]) { if(col[e1[i].u] != col[e1[i].v] ) { ans.push_back(i); } } else ans.push_back(i); } int size = ans.size(); sort(ans.begin(),ans.end()); printf("%d ",size); for(int i = 0; i < size ; i++ ) { printf("%d",ans[i]); if(i != size - 1 ) printf(" "); else printf(" "); } } int main() { int cas; cas = 0 ; while(scanf("%d%d",&n,&m)!=EOF) { cnt1 = 1 ; int u,v; double w; double L,R; L = eps; for(int i = 1 ; i <= m; i ++ ) { scanf("%d%d%lf",&u,&v,&w); e1[i].init(u,v,w); R = max(R , w); } double mid ; for(int i = 0 ; i < 101 ; i ++ ) // while( R - L > eps ) { mid = ( L + R ) / 2.0; double tmp = judge(mid); if(sgn(tmp ) > 0 ) // 判断 L = mid + eps; else R = mid; } if(cas != 0 ) printf(" "); cas ++ ; judge(L); getst(L); } return 0; }