poj2125最小点权覆盖+找一个割集

Destroying The Graph

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8503		Accepted: 2753		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 6
1 2 3
4 2 1
1 2
1 1
3 2
1 2
3 1
2 3

Sample Output

5
3
1 +
2 -
2 +

主要是找割边。
有构造出来的图知道这个是个二部图加两个源点汇点，二部图之间的连边不可能是割边（INF），所以就dfs(S)然后用vis标记，那么vis[S]一定是1，并且vis[T]一定是0.因为S，T不可能在一个集合里
那么从源点处找一下和它相连的边，看vis[]是不是0，是的话就是割边。
然后从汇点处找一下和它相连的边，看vis[]是不是1，是的话就是割边。
因为是个二部图所以不用dfs直接找一次就可以了

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=;
const int M=;
const int INF=1e9+;
int head[N],tot,S,T;
int q[N],dis[N],n,m,Q;
bool vis[N];
struct node
{
    int next,v,w;
} e[M<<];
void add(int u,int v,int w)
{
    e[tot].v=v;
    e[tot].w=w;
    e[tot].next=head[u];
    head[u]=tot++;
}
bool bfs()
{
    memset(dis,-,sizeof(dis));
    dis[S]=;
    int l=,r=;
    q[r++]=S;
    while(l<r)
    {
        int u=q[l++];
        for(int i=head[u]; ~i; i=e[i].next)
        {
            int v=e[i].v;
            if(dis[v]==-&&e[i].w>)
            {
                q[r++]=v;
                dis[v]=dis[u]+;
                if(v==T)  return true;
            }
        }
    }
    return false;
}
int dfs(int s,int low)
{
    if(s==T||!low) return low;
    int ans=low,a;
    for(int i=head[s]; ~i; i=e[i].next)
    {
        if(e[i].w>&&dis[e[i].v]==dis[s]+&&(a=dfs(e[i].v,min(e[i].w,ans))))
        {
            e[i].w-=a;
            e[i^].w+=a;
            ans-=a;
            if(!ans) return low;
        }
    }
    if(low==ans) dis[s]=-;
    return low-ans;
}
void dfs(int u){
    vis[u]=;
    for(int i=head[u];~i;i=e[i].next) if(!vis[e[i].v]&&e[i].w) dfs(e[i].v);
}
int a[N],b[N];
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        S=,T=*n+;
        int x,f,t;
        memset(head,-,sizeof(head));
        tot=;
        for(int i=; i<=n; ++i)
        {
            scanf("%d",&x);
            add(S,i,x);
            add(i,S,);
        }
        for(int i=; i<=n; ++i)
        {
            scanf("%d",&x);
            add(i+n,T,x);
            add(T,i+n,);
        }
        while(m--)
        {
            scanf("%d%d",&f,&t);
            add(t,f+n,INF);
            add(f+n,t,);
        }
        int ans=;
        while(bfs()) ans+=dfs(S,INF);
        printf("%d\n",ans);
        dfs(S);
        int ct1=,ct2=;
        for(int i=head[S];~i;i=e[i].next) if(!vis[e[i].v]) a[ct1++]=e[i].v;
        for(int i=head[T];~i;i=e[i].next) if(vis[e[i].v]) b[ct2++]=e[i].v-n;
        printf("%d\n",ct1+ct2);
        for(int i=;i<ct1;++i) printf("%d +\n",a[i]);
        for(int i=;i<ct2;++i) printf("%d -\n",b[i]);

    }
}