Language:
Destroying The Graph
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 7469   Accepted: 2381   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 +

一道单纯的最小点权覆盖，但是本小白还是不会写，看了题解，再回头看，突然发现输出路径那一块还是很好理解的，代码如下：

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
#include<climits>
#define MAXE 220*220
#define MAXP 220
#define Max(a,b) a>b?a:b
#define Min(a,b) a<b?a:b
using namespace std;
struct Edge
{
    int s,t,f,next;
} edge[MAXE];
int head[MAXP];
int cur[MAXP];
int pre[MAXP];
int stack[MAXE];
int sign[MAXE];
int ent;
int sum;
int n,m,s,t,supers,supert;
int num;
void add(int start,int last,int f)
{
    edge[ent].s=start;
    edge[ent].t=last;
    edge[ent].f=f;
    edge[ent].next=head[start];
    head[start]=ent++;
    edge[ent].s=last;
    edge[ent].t=start;
    edge[ent].f=0;
    edge[ent].next=head[last];
    head[last]=ent++;
}
bool bfs(int S,int T)
{
    memset(pre,-1,sizeof(pre));
    pre[S]=0;
    queue<int>q;
    q.push(S);
    while(!q.empty())
    {
        int temp=q.front();
        q.pop();
        for(int i=head[temp]; i!=-1; i=edge[i].next)
        {
            int temp2=edge[i].t;
            if(pre[temp2]==-1&&edge[i].f)
            {
                pre[temp2]=pre[temp]+1;
                q.push(temp2);
            }
        }
    }
    return pre[T]!=-1;
}
int dinic(int start,int last)
{
    int flow=0,now;
    while(bfs(start,last))
    {
        int top=0;
        memcpy(cur,head,sizeof(head));
        int u=start;
        while(1)
        {
            if(u==last)//如果找到终点结束对中间路径进行处理并计算出该流
            {
                int minn=INT_MAX;
                for(int i=0; i<top; i++)
                {
                    if(minn>edge[stack[i]].f)
                    {
                        minn=edge[stack[i]].f;
                        now=i;
                    }
                }
                flow+=minn;
                for(int i=0; i<top; i++)
                {
                    edge[stack[i]].f-=minn;
                    edge[stack[i]^1].f+=minn;
                }
                top=now;
                u=edge[stack[top]].s;
            }
            for(int i=cur[u]; i!=-1; cur[u]=i=edge[i].next) //找出从u点出发能到的边
                if(edge[i].f&&pre[edge[i].t]==pre[u]+1)
                    break;
            if(cur[u]==-1)//如果从该点未找到可行边，将该点标记并回溯
            {
                if(top==0)break;
                pre[u]=-1;
                u=edge[stack[--top]].s;
            }
            else//如果找到了继续运行
            {
                stack[top++]=cur[u];
                u=edge[cur[u]].t;
            }
        }
    }
    return flow;
}
void dfs(int S)
{
    sign[S]=1;
    for(int i=head[S]; i!=-1; i=edge[i].next)
    {
        if(edge[i].f>0&&!sign[edge[i].t])//如果与源相连的边未能满流，则说明不该删除那一点的出边，同时应该输出与那一点相连的同时与汇满流的点的所有入边。同时因为如果与源相连的变边满流，则直接不会因为这一点遍历到与这一点相连的边。

        {
            dfs(edge[i].t);
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(head,-1,sizeof(head));
        memset(sign,0,sizeof(sign));
        ent=0;
        s=0;
        t=2*n+1;
        int cost;
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&cost);
            add(i+n,t,cost);
        }
        int temp=ent;
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&cost);
            add(s,i,cost);
        }
        int temp2=ent;
        for(int i=1; i<=m; i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            add(u,v+n,INT_MAX);
        }
        printf("%d\n",dinic(s,t));
        dfs(s);int cnt=0;
        for(int i=1; i<=2*n; i++)
        {
            if(!sign[i]&&i<=n)cnt++;
            else if(sign[i]&&i>n)cnt++;
        }
        printf("%d\n",cnt);
        for(int i=1; i<=2*n; i++)
        {
            if(!sign[i]&&i<=n)printf("%d -\n",i);
            else if(sign[i]&&i>n)printf("%d +\n",i-n);
        }
    }
    return 0;
}