【POJ】【2125】Destroying the Graph

网络流/二分图最小点权覆盖

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　　果然还是应该先看下胡伯涛的论文……

　　orz proverbs

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。

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。。

　　汗……输出方案我WA了N次T_T，直接从S进行dfs，对于左边的点，如果走不到则表明 s->i 这条边被割掉了，对于右边的点，如果走的到则表明 i+n->t 这条边被割掉了，因为如果没割掉就直接从这个点走到t了……唉我一开始居然没想到

Source Code
Problem: 2125        User: sdfzyhy
Memory: 848K        Time: 79MS
Language: G++        Result: Accepted

    Source Code

    //BZOJ 2125
    #include<vector>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    #define rep(i,n) for(int i=0;i<n;++i)
    #define F(i,j,n) for(int i=j;i<=n;++i)
    #define D(i,j,n) for(int i=j;i>=n;--i)
    #define fore(i,x) for(int i=head[x];i;i=next[i])
    #define pb push_back
    using namespace std;
    inline int getint(){
        int v=0,sign=1; char ch=getchar();
        while(ch<'0'||ch>'9'){ if (ch=='-') sign=-1; ch=getchar();}
        while(ch>='0'&&ch<='9'){ v=v*10+ch-'0'; ch=getchar();}
        return v*sign;
    }
    const int N=310,M=20010,INF=~0u>>2;
    typedef long long LL;
    /******************tamplate*********************/

    struct edge{
        int from,to,v;
    };
    int n,m;
    struct Net{
        edge E[M];
        int head[N],next[M],cnt;
        void add(int x,int y,int z){
            E[++cnt]=(edge){x,y,z};
            next[cnt]=head[x]; head[x]=cnt;
            E[++cnt]=(edge){y,x,0};
            next[cnt]=head[y]; head[y]=cnt;
        }
        int s,t,d[N],cur[N],Q[N];
        void init(){
            n=getint(); m=getint();
            s=0; t=n*2+1; cnt=1;
            int x,y;
            F(i,1,n){
                x=getint();
                add(i+n,t,x);
            }
            F(i,1,n){
                x=getint();
                add(s,i,x);
            }
            F(i,1,m){
                x=getint(); y=getint();
                add(x,y+n,INF);
            }
        }
        bool mklevel(){
            memset(d,-1,sizeof d);
            d[s]=0;
            int l=0,r=-1;
            Q[++r]=s;
            while(l<=r){
                int x=Q[l++];
                fore(i,x){
                    edge&e=E[i];
                    if (d[e.to]==-1 && e.v>0){
                        d[e.to]=d[x]+1;
                        Q[++r]=e.to;
                    }
                }
            }
            return d[t]!=-1;
        }
        int dfs(int x,int a){
            if (x==t) return a;
            int flow=0;
            for(int &i=cur[x];i && flow<a;i=next[i]){
                edge&e=E[i];
                if (!e.v || d[e.to]!=d[x]+1) continue;
                int f=dfs(e.to,min(a-flow,e.v));
                if (f>0){
                    flow+=f;
                    e.v-=f;
                    E[i^1].v+=f;
                }
            }
            if (!flow) d[x]=-1;
            return flow;
        }
        int Dinic(){
            int flow=0;
            while(mklevel()){
                F(i,s,t) cur[i]=head[i];
                flow+=dfs(s,INF);
            }
            return flow;
        }
        bool vis[N];
        void dfs1(int x){
            if (vis[x]) return;
            vis[x]=1;
            for(int i=head[x];i;i=next[i])
                if (E[i].v) dfs1(E[i].to);
        }
        void solve(){
            printf("%d
",Dinic());
            int num=0;
            memset(vis,0,sizeof vis);
            dfs1(s);
            F(i,1,n) num+=(!vis[i])+vis[i+n];
            printf("%d
",num);
            F(i,1,n){
                if (!vis[i]) printf("%d -
",i);
                if (vis[i+n]) printf("%d +
",i);
            }
        }
    }G1;

    int main(){
    #ifndef ONLINE_JUDGE
        freopen("2125.in","r",stdin);
        freopen("2125.out","w",stdout);
    #endif
        G1.init();
        G1.solve();
        return 0;
    }

Destroying The Graph

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7511		Accepted: 2399		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 +

Source

Northeastern Europe 2003, Northern Subregion

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