洛谷P4014 分配问题（费用流）

传送门

 

可以把原图看做一个二分图，人在左边，任务在右边，求一个带权的最大和最小完美匹配

然而我并不会二分图做法，所以只好直接用费用流套进去，求一个最小费用最大流和最大费用最大流即可

  1 //minamoto
  2 #include
  3 #include
  4 #include
  5 #include
  6 #define inf 0x3f3f3f3f
  7 using namespace std;
  8 #define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
  9 char buf[1<<21],*p1=buf,*p2=buf;
 10 inline int read(){
 11     #define num ch-'0'
 12     char ch;bool flag=0;int res;
 13     while(!isdigit(ch=getc()))
 14     (ch=='-')&&(flag=true);
 15     for(res=num;isdigit(ch=getc());res=res*10+num);
 16     (flag)&&(res=-res);
 17     #undef num
 18     return res;
 19 }
 20 const int N=205,M=25005;
 21 int ver[M],Next[M],head[N],edge[M],flow[M],tot=1;
 22 int dis[N],disf[N],n,s,t,ans,Pre[N],last[N],vis[N],x;
 23 queue<int> q;
 24 inline void add(int u,int v,int e,int f){
 25     ver[++tot]=v,Next[tot]=head[u],head[u]=tot,edge[tot]=e,flow[tot]=f;
 26     ver[++tot]=u,Next[tot]=head[v],head[v]=tot,edge[tot]=-e,flow[tot]=0;
 27 }
 28 bool spfa_min(){
 29     memset(dis,0x3f,sizeof(dis));
 30     memset(vis,0,sizeof(vis));
 31     memset(disf,0x3f,sizeof(disf));
 32     q.push(s),dis[s]=0,vis[s]=1,Pre[t]=-1;
 33     while(!q.empty()){
 34         int u=q.front();q.pop();vis[u]=0;
 35         for(int i=head[u];i;i=Next[i]){
 36             int v=ver[i];
 37             if(flow[i]>0&&dis[v]>dis[u]+edge[i]){
 38                 dis[v]=dis[u]+edge[i],Pre[v]=u;
 39                 last[v]=i,disf[v]=min(disf[u],flow[i]);
 40                 if(!vis[v]) vis[v]=1,q.push(v);
 41             }
 42         }
 43     }
 44     return Pre[t]!=-1;
 45 }
 46 int dinic_min(){
 47     int maxflow=0,mincost=0;
 48     while(spfa_min()){
 49         int u=t;
 50         maxflow+=disf[t],mincost+=disf[t]*dis[t];
 51         while(u!=s){
 52             flow[last[u]]-=disf[t];
 53             flow[last[u]^1]+=disf[t];
 54             u=Pre[u];
 55         }
 56     }
 57     return mincost;
 58 }
 59 bool spfa_max(){
 60     memset(dis,0xef,sizeof(dis));
 61     memset(vis,0,sizeof(vis));
 62     memset(disf,0x3f,sizeof(disf));
 63     q.push(s),dis[s]=0,vis[s]=1,Pre[t]=-1;
 64     while(!q.empty()){
 65         int u=q.front();q.pop();vis[u]=0;
 66         for(int i=head[u];i;i=Next[i]){
 67             int v=ver[i];
 68             if(flow[i]>0&&dis[v]edge[i]){
 69                 dis[v]=dis[u]+edge[i],Pre[v]=u;
 70                 last[v]=i,disf[v]=min(disf[u],flow[i]);
 71                 if(!vis[v]) vis[v]=1,q.push(v);
 72             }
 73         }
 74     }
 75     return Pre[t]!=-1;
 76 }
 77 int dinic_max(){
 78     int maxflow=0,maxcost=0;
 79     while(spfa_max()){
 80         int u=t;
 81         maxflow+=disf[t],maxcost+=disf[t]*dis[t];
 82         while(u!=s){
 83             flow[last[u]]-=disf[t];
 84             flow[last[u]^1]+=disf[t];
 85             u=Pre[u];
 86         }
 87     }
 88     return maxcost;
 89 }
 90 int main(){
 91     n=read();
 92     for(int i=1;i<=n;++i)
 93     for(int j=1;j<=n;++j)
 94     x=read(),add(i,j+n,x,1);
 95     s=0,t=n+n+1;
 96     for(int i=1;i<=n;++i) add(s,i,0,1),add(i+n,t,0,1);
 97     printf("%d\n",dinic_min());
 98     for(int i=2;i<=tot;i+=2) flow[i]+=flow[i^1],flow[i^1]=0;
 99     printf("%d\n",dinic_max());
100     return 0;
101 }

 

转载于:https://www.cnblogs.com/bztMinamoto/p/9501411.html