題目連結

求一個割集 使得割集的平均邊權最小

具體的解答 請看Amber論文《最小割模型在資訊學競賽中的應用》

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 100010
#define MAXM 50100
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

struct Edge
{
    int to,next;
    double flow,cap;
}edge[MAXM];
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
void init()
{
    tol=0;
    MEM(head,-1);
}
void addedge(int u,int v,double w,double rw)
{
    edge[tol].to=v; edge[tol].cap=w; edge[tol].next=head[u];
    edge[tol].flow=0; head[u]=tol++;
    edge[tol].to=u; edge[tol].cap=rw; edge[tol].next=head[v];
    edge[tol].flow=0; head[v]=tol++;
}
double sap(int start,int ed,int N)
{
    MEM(gap,0);
    MEM(dep,0);
    memcpy(cur,head,sizeof(head));
//    for(int i=start;i<=ed;i++)
//        cur[i]=head[i];
    int u=start;
    pre[u]=-1;
    gap[0]=N;
    double ans=0;
    while(dep[start]<N)
    {
        if(u==ed)
        {
            double Min=INF*1.0;
            for(int i=pre[u];i!=-1;i=pre[edge[i^1].to])
                if(Min>edge[i].cap-edge[i].flow)
                    Min=edge[i].cap-edge[i].flow;
            for(int i=pre[u];i!=-1;i=pre[edge[i^1].to])
            {
                edge[i].flow+=Min;
                edge[i^1].flow-=Min;
            }
            u=start;
            ans+=Min;
            continue;
        }
        bool flag=0;
        int v;
        for(int i=cur[u];i!=-1;i=edge[i].next)
        {
            v=edge[i].to;
            if((edge[i].cap-edge[i].flow)>eps&&dep[v]+1==dep[u])
            {
                flag=1;
                cur[u]=pre[v]=i;
                break;
            }
        }
        if(flag)
        {
            u=v; continue;
        }
        int Min=N;
        for(int i=head[u];i!=-1;i=edge[i].next)
            if((edge[i].cap-edge[i].flow)>eps&&dep[edge[i].to]<Min)
        {
            Min=dep[edge[i].to];
            cur[u]=i;
        }
        gap[dep[u]]--;
        if(!gap[dep[u]]) return ans;
        dep[u]=Min+1;
        gap[dep[u]]++;
        if(u!=start) u=edge[pre[u]^1].to;
    }
    return ans;
}
struct point
{
    int u,v;
    double w;
}po[MAXM];
int vis[MAXN];
void dfs(int u)
{
    vis[u]=1;
    for(int i=head[u];i!=-1;i=edge[i].next)
        if(edge[i].cap-edge[i].flow>eps&&!vis[edge[i].to])
            dfs(edge[i].to);
}


int main()
{
//    fread;
    int n,m;
    int cs=0;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        double st=10000010,ed=0.0;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%lf",&po[i].u,&po[i].v,&po[i].w);
            st=min(st,po[i].w);
            ed=max(ed,po[i].w);
        }
//        double st=0.0,ed=10000010;
        double mid;
        double flow;
        while(ed-st>eps)
        {
            init();
            mid=(ed+st)/2.0;
            flow=0;
            for(int i=0;i<m;i++)
            {
                if(po[i].w<=mid) flow+=po[i].w-mid;
                else addedge(po[i].u,po[i].v,po[i].w-mid,po[i].w-mid);
            }
            flow+=sap(1,n,n);
            if(flow>0) st=mid;
            else ed=mid;
        }
        init();
        for(int i=0;i<m;i++)
        {
            if(po[i].w<=ed) continue;
            addedge(po[i].u,po[i].v,po[i].w-ed,po[i].w-ed);
        }
        sap(1,n,n);
        MEM(vis,0);
        dfs(1);
        vector<int> ans;
        for(int i=0;i<m;i++)
        {
            if((po[i].w<=ed)||(vis[po[i].u]&&!vis[po[i].v])||(!vis[po[i].u]&&vis[po[i].v]))
                ans.push_back(i+1);
        }
        int num=ans.size();
        if(cs) puts("");
        else cs=1;
        printf("%d\n",num);
        for(int i=0;i<num;i++)
        {
            if(i) printf(" ");
            printf("%d",ans[i]);
        }
        printf("\n");
        ans.clear();
    }
    return 0;
}