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POJ 2125 Destroying the Graph 二分图最小点权覆盖
Destroying The Graph
| Time Limit: 2000MS | Memory Limit: 65536K | |||
| Total Submissions: 8198 | Accepted: 2635 | Special Judge | ||
Description
Alice and Bob play the following game. First, Alice draws some directed graph with N vertices and M arcs. After that Bob tries to destroy it. In a move he may take any vertex of the graph and remove either all arcs incoming into this vertex, or all arcs outgoing from this vertex.
Alice assigns two costs to each vertex: Wi+ and Wi-. If Bob removes all arcs incoming into the i-th vertex he pays Wi+ dollars to Alice, and if he removes outgoing arcs he pays Wi- dollars.
Find out what minimal sum Bob needs to remove all arcs from the graph.
Alice assigns two costs to each vertex: Wi+ and Wi-. If Bob removes all arcs incoming into the i-th vertex he pays Wi+ dollars to Alice, and if he removes outgoing arcs he pays Wi- dollars.
Find out what minimal sum Bob needs to remove all arcs from the graph.
Input
Input file describes the graph Alice has drawn. The first line of the input file contains N and M (1 <= N <= 100, 1 <= M <= 5000). The second line contains N integer numbers specifying Wi+. The third line defines Wi- in a similar way. All costs are positive and do not exceed 106 . Each of the following M lines contains two integers describing the corresponding arc of the graph. Graph may contain loops and parallel arcs.
Output
On the first line of the output file print W --- the minimal sum Bob must have to remove all arcs from the graph. On the second line print K --- the number of moves Bob needs to do it. After that print K lines that describe Bob‘s moves. Each line must first contain the number of the vertex and then ‘+‘ or ‘-‘ character, separated by one space. Character ‘+‘ means that Bob removes all arcs incoming into the specified vertex and ‘-‘ that Bob removes all arcs outgoing from the specified vertex.
Sample Input
3 61 2 34 2 11 21 13 21 23 12 3
Sample Output
531 +2 -2 +
Source
Northeastern Europe 2003, Northern Subregion
【题意】：
N个点M条边的有向图，给出如下两种操作。
删除点i的所有出边，代价是Ai。
删除点j的所有入边，代价是Bj。
求最后删除图中所有的边的最小代价。
其实就是二分图最小点权覆盖。
定义：从x或者y集合中选取一些点，使这些点覆盖所有的边，并且选出来的点的权值尽可能小。
//最小点权覆盖就是求最小割（证明可参考胡伯涛论文“最小割模型在信息学竞赛中的应用”）。
【题解】：
拆点。n个点拆成2n个点（左右各n个，i与(i+n)对应，之间连容量INF的边），S和i连容量为Ai的边，(i+n)与T之间连容量为Bi的边，求最小割即可
这样做为什么对呢？
当一条边存在的条件就是网络中还存在从S到T的非满流边！
方案输出不多说。
//输出方案WA到挺的代码 #include<cstdio>#include<cstring>#include<algorithm>#define R register#define inf 0x3f3f3f3fusing namespace std;int read(){ R int x=0;bool f=1; R char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=0;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=(x<<3)+(x<<1)+ch-‘0‘;ch=getchar();} return f?x:-x;}const int N=1e5+10;struct node{ int v,next,cap,flow;}e[N<<2];int tot=1;struct data{ int x,op,val; bool operator <(const data &a)const{ return val==a.val?x<a.x:val<a.val; }}record[N];int n,m,cs,cc,S,T,a[N],b[N],cur[N],head[N],dis[N],q[N*2];bool mark[N];void add(int x,int y,int z){ e[++tot].v=y;e[tot].next=head[x];e[tot].cap=z;e[tot].flow=0;head[x]=tot; e[++tot].v=x;e[tot].next=head[y];e[tot].cap=0;e[tot].flow=0;head[y]=tot;}bool bfs(){ int h=0,t=1; memset(dis,-1,sizeof(dis)); dis[S]=0;q[1]=S; while(h!=t){ int x=q[++h]; for(int i=head[x];i;i=e[i].next){ int v=e[i].v; if(dis[v]==-1&&e[i].cap>e[i].flow){ dis[v]=dis[x]+1; q[++t]=v; } } } return dis[T]!=-1;}int dfs(int x,int f){ if(x==T||!f) return f; int used=0,f1; for(int &i=cur[x];i;i=e[i].next){ if(dis[x]+1==dis[e[i].v]&&(f1=dfs(e[i].v,min(f,e[i].cap-e[i].flow)))>0){ e[i].flow+=f1;e[i^1].flow-=f1; used+=f1;f-=f1; if(!f) break; } } return used;}int dinic(){ int ans=0; while(bfs()){ for(int i=S;i<=T;i++) cur[i]=head[i]; ans+=dfs(S,0x7fffffff); } return ans;}void dfs_cut(int x){ if(x==T) return ; mark[x]=1; for(int i=head[x];i;i=e[i].next){ int v=e[i].v,val,op; if(!mark[v]){ if(e[i].cap==e[i].flow){ if(x!=S){ if(x>n) op=0,val=b[x-n]; else op=1,val=a[x]; record[++cc].x=x;record[cc].op=op;record[cc].val=val; } if(v!=T){ if(v>n) op=0,val=b[v-n]; else op=1,val=a[v]; record[++cc].x=v;record[cc].op=op;record[cc].val=val; } } dfs_cut(v); } }}int main(){ n=read();m=read(); S=0;T=n<<1|1; for(int i=1;i<=n;i++) a[i]=read(),add(S,i,a[i]); for(int i=1;i<=n;i++) b[i]=read(),add(i+n,T,b[i]); for(int i=1,x,y;i<=m;i++) x=read(),y=read(),add(x,y+n,inf); printf("%d\n",dinic()); dfs_cut(S); for(int i=1;i<=n;i++){ if(!mark[i]) cs++; if(mark[i+n]) cs++; } printf("%d\n",cs); for(int i=1;i<=cc;i++) if(record[i].x>n) record[i].x-=n; sort(record+1,record+cc+1); for(int i=1;i<=cs;i++){ int &x=record[i].x,&y=record[i].op; printf("%d ",x);putchar(y?‘+‘:‘-‘);printf("\n"); } return 0;}//from zjk‘s AC code#include<cstdio>#include<iostream>#define N 210#define M 5010#define inf 1000000000using namespace std;int a[N],b[N],head[N],dis[N],q[N],vis[N],n,m,cnt=1,S,T;struct node{ int v,f,pre;}e[M*2];void add(int u,int v,int f){ e[++cnt].v=v;e[cnt].f=f;e[cnt].pre=head[u];head[u]=cnt; e[++cnt].v=u;e[cnt].f=0;e[cnt].pre=head[v];head[v]=cnt;}bool bfs(){ for(int i=1;i<=T;i++)dis[i]=inf; int h=0,t=1;q[1]=S;dis[S]=0; while(h<t){ int u=q[++h]; for(int i=head[u];i;i=e[i].pre){ int v=e[i].v; if(e[i].f&&dis[u]+1<dis[v]){ dis[v]=dis[u]+1; if(v==T)return true; q[++t]=v; } } } if(dis[T]==inf)return false; return true;}int dinic(int now,int f){ if(now==T)return f; int rest=f; for(int i=head[now];i;i=e[i].pre){ int v=e[i].v; if(e[i].f&&dis[v]==dis[now]+1){ int t=dinic(v,min(rest,e[i].f)); if(!t)dis[v]=0; e[i].f-=t; e[i^1].f+=t; rest-=t; } } return f-rest;}void dfs(int x){ vis[x]=1; for(int i=head[x];i;i=e[i].pre){ if(!e[i].f||vis[e[i].v])continue; dfs(e[i].v); }}int main(){ scanf("%d%d",&n,&m); S=0,T=2*n+1; for(int i=1;i<=n;i++){ scanf("%d",&a[i]); add(i+n,T,a[i]); } for(int i=1;i<=n;i++){ scanf("%d",&b[i]); add(S,i,b[i]); } for(int i=1;i<=m;i++){ int x,y;scanf("%d%d",&x,&y); add(x,y+n,inf); } int min_cnt=0,p=0,pin=0,pout=0; while(bfs()) min_cnt+=dinic(S,inf); printf("%d\n",min_cnt); dfs(S); for(int i=1;i<=n;i++){ if(!vis[i])p++; if(vis[i+n])p++; } printf("%d\n",p); for(int i=1;i<=n;i++){ if(!vis[i])printf("%d -\n",i); if(vis[i+n])printf("%d +\n",i); } return 0;}
POJ 2125 Destroying the Graph 二分图最小点权覆盖
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