Poj 3281 (max stream + match + split point)

Question link:Http://poj.org/problem? Id = 3281

Theme: There are some cows, a pile of food, and a pile of drinks. If a cow eats a food and drinks a drink, it is satisfying. In addition, the ox is fond of certain foods and drinks and asks how many cows are satisfied at most.

Solutions:

There is no matching maximum flow question for the fee.

At the beginning, I thought so. S-> ox-> food-> drinks-> T, and cap are all 1, aha. It's easy.

This is wrong. Because the ox is picky, suppose the OX 1 prefers food 1 rather than drink 1, then the food 1 and drink 1 are not connected?

Don't connect. What about other cows?

What should I do with niu 1?

You cannot create a chart in this way. Then we changed it to s-> food-> ox-> drinks-> T, which solved the problem of ambiguous relationship between food and drinks.

But the new problem comes again. Suppose there is food 1> OX 1> drink 1, then there will be food 2> OX 1> drink 2, in this way, a cow swallowed up more food and drinks.

The solution is to split the point and insert the ox into the ox-> ox, Cap = 1, so that a cow will only be used once.

The final scheme for creating images: S-> food-> ox-> drinks-> T.

 

#include "cstdio"#include "vector"#include "cstring"#include "queue"using namespace std;#define maxn 405#define inf 100000000struct Edge{    int from,to,cap,flow;    Edge(int FROM,int TO,int CAP,int FLOW):from(FROM),to(TO),cap(CAP),flow(FLOW) {}};int d[maxn],p[maxn],gap[maxn],cur[maxn];bool vis[maxn];vector<int> G[maxn],food[105],drink[105];vector<Edge> edges;void addedge(int from,int to,int cap){    edges.push_back(Edge(from,to,cap,0));    edges.push_back(Edge(to,from,0,0));    int m=edges.size();    G[from].push_back(m-2);    G[to].push_back(m-1);}void bfs(int s,int t){    memset(vis,false,sizeof(vis));    memset(d,0,sizeof(d));    memset(p,0,sizeof(p));    d[t]=0;vis[t]=true;    queue<int> Q;Q.push(t);    while(!Q.empty())    {        int u=Q.front();Q.pop();        for(int v=0;v<G[u].size();v++)        {            Edge e=edges[G[u][v]^1];            if(!vis[e.from]&&e.cap>e.flow)            {                vis[e.from]=true;                d[e.from]=d[u]+1;                Q.push(e.from);            }        }    }}int augment(int s,int t){    int x=t,a=inf;    while(x!=s)    {        Edge e=edges[p[x]];        a=min(a,e.cap-e.flow);        x=e.from;    }    x=t;    while(x!=s)    {        edges[p[x]].flow+=a;        edges[p[x]^1].flow-=a;        x=edges[p[x]].from;    }    return a;}int maxflow(int s,int t){    int flow=0,u=s;    bfs(s,t);    memset(gap,0,sizeof(gap));    memset(cur,0,sizeof(cur));    for(int i=0;i<=t;i++) gap[d[i]]++;    while(d[s]<t+1)    {        if(u==t)        {            flow+=augment(s,t);            u=s;        }        bool flag=false;        for(int v=cur[u];v<G[u].size();v++) //Advance        {            Edge e=edges[G[u][v]];            if(e.cap>e.flow&&d[u]==d[e.to]+1)            {                flag=true;                p[e.to]=G[u][v];                cur[u]=v;                u=e.to;                break;            }        }        if(!flag) //Retreat        {            int m=t+1;            for(int v=0;v<G[u].size();v++)            {                Edge e=edges[G[u][v]];                if(e.cap>e.flow) m=min(m,d[e.to]);            }            if(--gap[d[u]]==0) break;            gap[d[u]=m+1]++;            cur[u]=0;            if(u!=s) u=edges[p[u]].from;        }    }    return flow;}int main(){    int N,M,K,f,d,t;    while(scanf("%d%d%d",&N,&M,&K)!=EOF)    {        for(int i=1;i<=M;i++) addedge(0,i,1); //S-food        for(int i=1;i<=K;i++) addedge(i+M+N*2,K+M+N*2+1,1); //drink-T        for(int i=1;i<=N;i++)        {            addedge(i+M,i+M+N,1); //cow-cow‘            scanf("%d%d",&f,&d);            for(int j=1; j<=f; j++)            {                scanf("%d",&t);                addedge(t,i+M,1); //food-cow            }            for(int j=1; j<=d; j++)            {                scanf("%d",&t);                addedge(i+M+N,t+M+2*N,1); //cow‘-drink            }        }        printf("%d\n",maxflow(0,2*N+M+K+1));    }}

 

13456393

Neopenx

3281

Accepted

320 k

16 Ms

C ++

3222b

00:28:49

Poj 3281 (max stream + match + split point)