【题目分析】
无源汇上下界可行流。
上下界网络流的问题可以参考这里。↓
http://www.cnblogs.com/kane0526/archive/2013/04/05/3001108.html
【代码】
#include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> //#include <map> #include <set> #include <queue> #include <string> #include <iostream> #include <algorithm> using namespace std; #define maxn 205 #define me 50005 #define inf 0x3f3f3f3f #define F(i,j,k) for (int i=j;i<=k;++i) #define D(i,j,k) for (int i=j;i>=k;--i) void Finout() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout); #endif } int Getint() { int x=0,f=1; char ch=getchar(); while (ch<‘0‘||ch>‘9‘) {if (ch==‘-‘) f=-1; ch=getchar();} while (ch>=‘0‘&&ch<=‘9‘) {x=x*10+ch-‘0‘; ch=getchar();} return x*f; } int h[me<<1],to[me<<1],ne[me<<1],fl[me<<1],en=0,S=0,T=me-1; int id[me<<1]; void add(int a,int b,int c,int ID) { // cout<<"add "<<a<<" "<<b<<" "<<c<<endl; to[en]=b; ne[en]=h[a]; fl[en]=c; id[en]=ID; h[a]=en++; to[en]=a; ne[en]=h[b]; fl[en]=0; id[en]=0; h[b]=en++; } int map[me]; bool tell() { queue <int> q; memset(map,-1,sizeof map); map[S]=0; while (!q.empty()) q.pop(); q.push(S); while (!q.empty()) { int x=q.front(); q.pop(); // cout<<"bfs"<<x<<endl; for (int i=h[x];i>=0;i=ne[i]) { // cout<<"to "<<to[i]<<endl; if (map[to[i]]==-1&&fl[i]>0) { map[to[i]]=map[x]+1; q.push(to[i]); } } } // cout<<"over"<<endl; if (map[T]!=-1) return true; return false; } int zeng(int k,int r) { if (k==T) return r; int ret=0; for (int i=h[k];i>=0&&ret<r;i=ne[i]) if (map[to[i]]==map[k]+1&&fl[i]>0) { int tmp=zeng(to[i],min(fl[i],r-ret)); ret+=tmp; fl[i]-=tmp; fl[i^1]+=tmp; } if (!ret) map[k]=-1; return ret; } int ans[me<<1],n,m,du[me<<1],dn[me<<1]; int main() { Finout(); while (scanf("%d%d",&n,&m)!=EOF) { memset(h,-1,sizeof h); memset(ans,0,sizeof ans); memset(du,0,sizeof du); memset(dn,0,sizeof dn); F(i,1,m) { int a,b,c,d; a=Getint();b=Getint();c=Getint();d=Getint(); add(a,b,d-c,i); du[a]-=c; du[b]+=c; dn[i]+=c; } F(i,1,n) { if (du[i]) add(S,i,du[i],0); if (du[i]<0) add(i,T,-du[i],0); } int now=0,tmp=0; while (tell()) while (tmp=zeng(S,inf)) now+=tmp; int flag=1; for (int i=h[S];i>=0;i=ne[i]) if (fl[i]>0) flag=0; if (!flag) printf("NO\n"); else { printf("YES\n"); F(i,0,en-1) ans[id[i]]=fl[i^1]; F(i,1,m) printf("%d\n",ans[i]+dn[i]); } } }
时间: 2024-10-24 15:23:22