题目链接:http://poj.org/problem?id=2125
题意:
思路:将每个点i拆成两个点xi,yi,分别表示两种操作,每条有向边<u,v>连接<xu,yv,INF>,s向每个点连边<s,xi,ai>,每个点向t连边<yi,t,bi>。
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <map>
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)<(y)?(x):(y))
#define abs(x) ((x)>=0?(x):-(x))
#define i64 long long
#define u32 unsigned int
#define u64 unsigned long long
#define clr(x,y) memset(x,y,sizeof(x))
#define CLR(x) x.clear()
#define ph(x) push(x)
#define pb(x) push_back(x)
#define Len(x) x.length()
#define SZ(x) x.size()
#define PI acos(-1.0)
#define sqr(x) ((x)*(x))
#define MP(x,y) make_pair(x,y)
#define EPS 1e-9
#define FOR0(i,x) for(i=0;i<x;i++)
#define FOR1(i,x) for(i=1;i<=x;i++)
#define FOR(i,a,b) for(i=a;i<=b;i++)
#define FORL0(i,a) for(i=a;i>=0;i--)
#define FORL1(i,a) for(i=a;i>=1;i--)
#define FORL(i,a,b)for(i=a;i>=b;i--)
using namespace std;
void RD(int &x){scanf("%d",&x);}
void RD(u32 &x){scanf("%u",&x);}
void RD(double &x){scanf("%lf",&x);}
void RD(int &x,int &y){scanf("%d%d",&x,&y);}
void RD(u32 &x,u32 &y){scanf("%u%u",&x,&y);}
void RD(double &x,double &y){scanf("%lf%lf",&x,&y);}
void RD(int &x,int &y,int &z){scanf("%d%d%d",&x,&y,&z);}
void RD(int &x,int &y,int &z,int &t){scanf("%d%d%d%d",&x,&y,&z,&t);}
void RD(u32 &x,u32 &y,u32 &z){scanf("%u%u%u",&x,&y,&z);}
void RD(double &x,double &y,double &z){scanf("%lf%lf%lf",&x,&y,&z);}
void RD(char &x){x=getchar();}
void RD(char *s){scanf("%s",s);}
void RD(string &s){cin>>s;}
void PR(int x) {printf("%d\n",x);}
void PR(int x,int y) {printf("%d %d\n",x,y);}
void PR(i64 x) {printf("%lld\n",x);}
void PR(u32 x) {printf("%u\n",x);}
void PR(double x) {printf("%.5lf\n",x);}
void PR(char x) {printf("%c\n",x);}
void PR(char *x) {printf("%s\n",x);}
void PR(string x) {cout<<x<<endl;}
struct Node
{
int u,v,cap,flow,next;
};
const int INF=1000000000;
Node edges[120005];
int n,m,s,t,head[1005],e;
int num[1005],h[1005],curedge[1005],pre[1005];
void add(int u,int v,double cap)
{
edges[e].u=u;
edges[e].v=v;
edges[e].cap=cap;
edges[e].flow=0.0;
edges[e].next=head[u];
head[u]=e++;
}
void Add(int u,int v,double cap)
{
add(u,v,cap);
add(v,u,0.0);
}
int Maxflow(int s,int t,int n)
{
int ans=0,i,k,x,d,u;
memset(num,0,sizeof(num));
memset(h,0,sizeof(h));
for(i=0;i<=n;i++) curedge[i]=head[i];
num[n]=n;u=s;
while(h[u]<n)
{
if(u==t)
{
d=INF+1;
for(i=s;i!=t;i=edges[curedge[i]].v) if(d>edges[curedge[i]].cap)
k=i,d=edges[curedge[i]].cap;
for(i=s;i!=t;i=edges[curedge[i]].v)
{
x=curedge[i];
edges[x].cap-=d;
edges[x].flow+=d;
edges[x^1].cap+=d;
edges[x^1].flow-=d;
}
ans+=d;u=k;
}
for(i=curedge[u];i!=-1;i=edges[i].next) if(edges[i].cap>0&&h[u]==h[edges[i].v]+1)
break;
if(i!=-1)
{
curedge[u]=i;
pre[edges[i].v]=u;
u=edges[i].v;
}
else
{
if(--num[h[u]]==0) break;
curedge[u]=head[u];
for(x=n,i=head[u];i!=-1;i=edges[i].next) if(edges[i].cap>0&&h[edges[i].v]<x)
x=h[edges[i].v];
h[u]=x+1;num[h[u]]++;
if(u!=s) u=pre[u];
}
}
return ans;
}
int visit[205];
void DFS(int u)
{
visit[u]=1;
int i;
for(i=head[u];i!=-1;i=edges[i].next)
{
if(!visit[edges[i].v]&&edges[i].cap>0)
{
DFS(edges[i].v);
}
}
}
int main()
{
RD(n,m);clr(head,-1); e=0; s=0;t=n+n+1;
int i,u,v,x;
FOR1(i,n) RD(x),Add(i+n,t,x);
FOR1(i,n) RD(x),Add(s,i,x);
FOR1(i,m) RD(u,v),Add(u,v+n,INF);
PR(Maxflow(s,t,t+1));
clr(visit,0); DFS(s);
vector<int> ans;
FOR1(i,2*n) if(visit[i]&&i>n||!visit[i]&&i<=n)
{
ans.push_back(i);
}
PR(SZ(ans));
FOR0(i,SZ(ans))
{
if(ans[i]<=n) printf("%d -\n",ans[i]);
else printf("%d +\n",ans[i]-n);
}
return 0;
}