Gym

题目链接：

https://odzkskevi.qnssl.com/19afdbfcc9ee8720dd81032e85ed2a5a?v=1506446502

题解：

二分匹配或网络流。

代码：

Dinic模板：

#include <cmath>#include <queue>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define met(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3ftypedef long long ll;const ll MAX_E=2500+10;//edgeconst ll MAX_P=2500+10;//pointstruct node{    ll to,next,w;}edge[MAX_E*MAX_E];ll head[MAX_P*10],cur[MAX_P*10];ll a[MAX_P*10],b[MAX_P*10];ll pos[MAX_P*10][3];ll dis[MAX_P*10];ll w[MAX_P*10];ll top;ll s,t;void init(){    top=1;    met(head,-1);    met(cur,-1);}void Add(ll u,ll v,ll w){    edge[++top].to=v;    edge[top].w=w;    edge[top].next=head[u];    head[u]=top;}void Add_edge(ll u,ll v,ll w){    Add(u,v,w);    Add(v,u,0);}ll bfs(){    queue<ll>q;    met(dis,0);    while(!q.empty())        q.pop();    dis[s]=1;    q.push(s);    while(!q.empty())    {        ll x=q.front();        q.pop();        for(ll i=head[x];i!=-1;i=edge[i].next)        {            ll y=edge[i].to;            if(!dis[y]&&edge[i].w)            {                dis[y]=dis[x]+1;                q.push(y);            }        }    }    return dis[t];}ll dfs(ll x,ll flow){    if(x==t||flow==0)        return flow;    ll ans=0;    for(ll &i=cur[x];i!=-1;i=edge[i].next)//    for(ll i=head[x];i!=-1;i=edge[i].next)    {        ll y=edge[i].to;        if(dis[y]==dis[x]+1&&edge[i].w)        {            ll a=dfs(y,min(edge[i].w,flow-ans));            edge[i].w-=a;            edge[i^1].w+=a;            ans+=a;            if(flow==ans)                break;        }    }    return ans;}ll Dinic(){    ll ans=0;    while(bfs())    {        memcpy(cur,head,sizeof(head));        ll temp=dfs(s,inf);//        while(ll temp=dfs(s,inf))            ans+=temp;    }    return ans;}int main(){    ll n;    while(scanf("%lld",&n)!=EOF)    {        met(a,0);met(b,0);        met(pos,0);met(w,0);        for(ll i=1;i<=n;i++)        {            scanf("%lld%lld",&a[i],&b[i]);            w[i*3-2]=a[i]+b[i];            w[i*3-1]=a[i]-b[i];            w[i*3]=a[i]*b[i];        }        sort(w+1,w+1+n*3);        ll len=unique(w+1,w+1+n*3)-w-1;        s=0,t=n+len+1;        init();        for(ll i=1;i<=n;i++)        {            Add_edge(s,i,1);            ll x=lower_bound(w+1,w+len+1,a[i]+b[i])-w;            Add_edge(i,x+n,1);            pos[i][0]=top;            x=lower_bound(w+1,w+len+1,a[i]-b[i])-w;            Add_edge(i,x+n,1);            pos[i][1]=top;            x=lower_bound(w+1,w+len+1,a[i]*b[i])-w;            Add_edge(i,x+n,1);            pos[i][2]=top;        }        for(ll i=1;i<=len;i++)            Add_edge(i+n,t,1);        ll ans=Dinic();        if(ans<n)            printf("impossible\n");        else        {            for(ll i=1;i<=n;i++)            {                if (edge[pos[i][0]].w)                    printf("%lld + %lld = %lld\n",a[i],b[i],a[i]+b[i]);                else if (edge[pos[i][1]].w)                    printf("%lld - %lld = %lld\n",a[i],b[i],a[i]-b[i]);                else                    printf("%lld * %lld = %lld\n",a[i],b[i],a[i]*b[i]);            }        }    }}