★ poj 2125 二分图的最小点权覆盖+输出解

Destroying The Graph

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 7104 Accepted: 2256 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.

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

分析：就是二分图的最小点权覆盖问题，输出解的话，只要dfs即可。源点S，汇点T，将其余每个点拆成两个点，i和i+n，S和每一个i点连一条边，T点与每个i+n点连一条边，跑一遍最大流即可，对于输出解的情况，只要从S点开始dfs，标记走到的点即可。

代码：

//Dinic算法，复杂度O(n^2m)#pragma comment(linker,"/STACK:102400000,102400000")#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>#include <vector>#include <string>#include <math.h>#include <queue>#include <stack>#include <map>#include <set>using namespace std;typedef long long ll;   //记得必要的时候改成无符号const int maxn=505;const int maxm=1000005;const int INF=1000000000;struct EdgeNode{    int from;    int to;    int flow;    int next;}edge[maxm];int head[maxn],cnt;void add(int x,int y,int z){    edge[cnt].from=x;edge[cnt].to=y;edge[cnt].flow=z;edge[cnt].next=head[x];head[x]=cnt++;    edge[cnt].from=y;edge[cnt].to=x;edge[cnt].flow=0;edge[cnt].next=head[y];head[y]=cnt++;    //printf("%d %d %d\n",x,y,z);}void init(){    cnt=0;    memset(head,-1,sizeof(head));}int S,T,n,m;int d[maxn];bool bfs(int S,int T){    int que[maxn],iq=0,top;    memset(d,0,sizeof(d));    d[S]=1;    que[iq++]=S;    for(int i=0;i<iq;i++)    {        top=que[i];        if(top==T)return true;        for(int k=head[top];k!=-1;k=edge[k].next)        {            if(!d[edge[k].to]&&edge[k].flow)            {                que[iq++]=edge[k].to;                d[edge[k].to]=d[top]+1;            }        }    }    return false;}int dfs(int now,int maxf,int T){    if(now==T)return maxf;    int ret=0,f;    for(int k=head[now];k!=-1;k=edge[k].next)    {        if(edge[k].flow&&d[edge[k].to]==d[now]+1)        {            f=dfs(edge[k].to,min(maxf-ret,edge[k].flow),T);            edge[k].flow-=f;            edge[k^1].flow+=f;            ret+=f;            if(ret==maxf)return ret;        }    }    return ret;}int Dinic(int S,int T){    int ans=0;    while(bfs(S,T))ans+=dfs(S,INF,T);    return ans;}int a[maxn],b[maxn];bool mark[3*maxn];void dfs_cut(int x){    mark[x]=1;    int y;    for(int i=head[x];i!=-1;i=edge[i].next)    {        y=edge[i].to;        if(edge[i].flow&&!mark[y])            dfs_cut(y);    }}int main(){    int i,x,y,cs;    while(~scanf("%d%d",&n,&m))    {        init(); S=0; T=2*n+1;        for(i=1;i<=n;i++){            scanf("%d",&a[i]);            add(i+n,T,a[i]);        }        for(i=1;i<=n;i++){            scanf("%d",&b[i]);            add(S,i,b[i]);        }        for(i=1;i<=m;i++){            scanf("%d%d",&x,&y);            add(x,y+n,INF);        }        printf("%d\n",Dinic(S,T));        memset(mark,false,sizeof(mark));        dfs_cut(S); cs=0;        for(i=1;i<=n;i++)if(!mark[i])cs++;        for(i=n+1;i<=2*n;i++)if(mark[i])cs++;        printf("%d\n",cs);        for(i=1;i<=n;i++)if(!mark[i])printf("%d -\n",i);        for(i=n+1;i<=2*n;i++)if(mark[i])printf("%d +\n",i-n);    }    return 0;}