【最大流，二分图匹配】【hdu2063】【过山车】

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题意：裸的求二分图匹配

建立一个源点连向一边所有的点容量为1;

另外一边点都连向汇点容量为1;

二分图的边容量也为1

源点汇点求一遍最大流即可

#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <ctime>#include <algorithm>#include <iostream>#include <sstream>#include <string>#define oo 0x13131313using namespace std;const int MAXN=2000+5;const int MAXM=10000+5;const int INF=0x3f3f3f3f;struct Edge{    int to,next,cap,flow;    void get(int a,int b,int c,int d)    {        to=a;next=b;cap=c;flow=d;    }}edge[MAXM];int tol;int head[MAXN];int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];void init(){    tol=0;    memset(head,-1,sizeof(head));}//单向图三个参数，无向图四个参数void addedge(int u,int v,int w,int rw=0){    edge[tol].get(v,head[u],w,0);head[u]=tol++;    edge[tol].get(u,head[v],rw,0);head[v]=tol++;}int sap(int start,int end,int N){    memset(gap,0,sizeof(gap));    memset(dep,0,sizeof(dep));    memcpy(cur,head,sizeof(head));    int u=start;    pre[u]=-1;    gap[0]=N;    int ans=0;    while(dep[start]<N)    {        if(u==end)        {            int Min=INF;            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=false;        int v;        for(int i=cur[u];i !=-1;i=edge[i].next)        {            v=edge[i].to;            if(edge[i].cap-edge[i].flow&&dep[v]+1==dep[u])            {                flag=true;                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&&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;}int KK,MM,NN;void input(){    int a,b;    for(int i=1;i<=KK;i++)    {        scanf("%d%d",&a,&b);        addedge(a,b+MM,1);    }}void solve(){    int ANS;    for(int i=1;i<=MM;i++)    {        addedge(MM+NN+1,i,1);    }    for(int i=MM+1;i<=MM+NN;i++)    {        addedge(i,MM+NN+2,1);    }    ANS=sap(MM+NN+1,MM+NN+2,MM+NN+2);    printf("%d\n",ANS);}void File(){    freopen("a.in","r",stdin);    freopen("a.out","w",stdout);}int main(){  //  File();    while(cin>>KK>>MM>>NN&&KK)    {        init();        input();        solve();    }}

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