ZOJ 2567Trade(有下界的最小流)

最新推荐文章于 2020-12-17 21:46:16 发布

转载最新推荐文章于 2020-12-17 21:46:16 发布 · 170 阅读

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原文链接：http://www.cnblogs.com/wangsouc/articles/4008619.html

本文探讨了在N*M的二分图中，求解最小边数使每个顶点至少被两条边覆盖的问题。采用dinic算法实现，首先满足所有顶点的下界条件，再通过最大流算法寻找额外的边，确保每个顶点至少被覆盖两次。

给定N * M的二分图，求最小的边数使得每个点都被至少覆盖2次。

有下界的最小流问题。

先把需要满足的下界跑满（即减去容量）然后再跑不需要的边的最大流。

dinic

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<vector>
#define LL int

using namespace std;

const int maxn = 1011;
const int maxm = 191010;
const LL inf = 10000000;




struct Dinic{
    int node, src, dest, ne;
    int head[maxn], work[maxn], dist[maxn],Q[maxn];
    LL flow[maxm];
    int pnt[maxm], nxt[maxm];


    void init(int _node, int _src, int _dest)
    {
        node = _node;
        src = _src;
        dest = _dest;
        memset(head, -1 ,sizeof(head));
        ne = 0;
    }
    void add(int u,int v, LL c1,LL c2)
    {
        pnt[ne] = v, flow[ne] = c1, nxt[ne] = head[u], head[u] = ne++;
        pnt[ne] = u, flow[ne] = c2, nxt[ne] = head[v], head[v] = ne++;
    }
    bool dinic_bfs(void)
    {
        int i,u,v,l,r=0;
        for(int i=0;i<node;i++)    dist[i]=-1;
        dist[Q[r++]=src]=0;
        for(l=0;l<r;l++)    for(i=head[u=Q[l]];i>=0;i=nxt[i])
            if(flow[i] && dist[v=pnt[i]]<0)
            {
                dist[Q[r++]=v]=dist[u]+1;
                if(v==dest)    return 1;
            }
        return 0;
    }
    LL dinic_dfs(int u,LL exp)
    {
        if(u==dest)    return exp;
        LL tmp;
        for(int &i=work[u],v;i>=0;i=nxt[i])
            if(flow[i] && dist[v=pnt[i]]==dist[u]+1 && (tmp=dinic_dfs(v,min(exp,flow[i])))>0)
            {
                flow[i]-=tmp;
                flow[i^1]+=tmp;
                return tmp;
            }
        return 0;
    }
    LL dinic_flow(void)
    {
        int i;
        LL res=0,delta;
        while(dinic_bfs())
        {
            for(i=0;i<node;i++)    work[i]=head[i];
            while(delta=dinic_dfs(src,inf))    res+=delta;
        }
        return res;
    }
}Flow;

int n, m, q;
int deg[606];
int main()
{
    while (scanf("%d%d%d",&n,&m,&q)==3)
    {
        Flow.init(n + m + 2, 0, n + m + 1);
        memset(deg, 0, sizeof(deg));
        for (int i = 1;i<=q;i++)
        {
            int u, v;
            scanf("%d%d",&u,&v);
            Flow.add(u, n + v, 1 ,0);
            deg[u]++;
            deg[v+ n]++;
        }
        int flag = 1;
        for (int i=1;i<=n + m;i++)
            if (deg[i] < 2){
                flag = 0;
                break;
            }
        if (flag == 0)
        {
            printf("-1\n\n");
        }else{
            for (int i=1;i<=n;i++)
                Flow.add(0, i, deg[i] - 2, 0);
            for (int j = 1;j<=m;j++)
                Flow.add(n + j, n + m + 1, deg[n + j] - 2, 0);
            int ret = Flow.dinic_flow();
            printf("%d\n",q - ret);
            vector<int>ans;
            ans.clear();
            for (int i=1;i<=q;i++)
            {
                int e = i * 2 - 2;
                if (Flow.flow[e] == 1){
                    ans.push_back(i);
                }
            }
            int sz  = ans.size();
            for (int i=0;i<sz;i++)
            {
                if (i == sz - 1)printf("%d\n",ans[i]);
                else printf("%d ",ans[i]);
            }
        }
    }
    return 0;
}

View Code

转载于:https://www.cnblogs.com/wangsouc/articles/4008619.html