loj115 无源汇有上下界可行流

link

题意&题解

code:

 #include<bits/stdc++.h>
 #define rep(i,x,y) for (int i=(x);i<=(y);i++)
 #define ll long long
 #define inf 1000000001
 #define y1 y1___
 using namespace std;
 char gc(){
     static char buf[],*p1=buf,*p2=buf;
     return p1==p2&&(p2=(p1=buf)+fread(buf,,,stdin),p1==p2)?EOF:*p1++;
 }
 #define gc getchar
 ll read(){
     char ch=gc();ll x=;int op=;
     for (;!isdigit(ch);ch=gc()) if (ch=='-') op=-;
     for (;isdigit(ch);ch=gc()) x=(x<<)+(x<<)+ch-'';
     return x*op;
 }
 #define N 205
 #define M 10205+N<<1
 int n,m,cnt=,s,t,head[N],d[N],vis[N],cur[N];
 struct edge{int to,nxt,d,c;}e[M];
 void adde(int x,int y,int d,int c){
     e[++cnt].to=y;e[cnt].nxt=head[x];head[x]=cnt;
     e[cnt].d=d;e[cnt].c=c;//d：原图下界；c：新图容量
 }
 void ins(int x,int y,int d,int z){
     adde(x,y,d,z);adde(y,x,d,);
 }
 bool bfs(){
     queue<int> q;q.push(s);
     rep (i,,t) vis[i]=-;vis[s]=;
     while (!q.empty()){
         int u=q.front();q.pop();
         for (int i=head[u];i;i=e[i].nxt){
             int v=e[i].to;
             if (e[i].c&&vis[v]==-) vis[v]=vis[u]+,q.push(v);
         }
     }
     return vis[t]!=-;
 }
 int dfs(int u,int flow){
     if (u==t) return flow;
     int w,used=;
     for (int &i=cur[u];i;i=e[i].nxt){
         int v=e[i].to;
         if (e[i].c&&vis[v]==vis[u]+){
             w=dfs(v,min(flow-used,e[i].c));
             e[i].c-=w,e[i^].c+=w,used+=w;
             if (used==flow) return used;
         }
     }
     if (!used) vis[u]=-;
     return used;
 }
 int dinic(){
     int ret=;
     while (bfs()){
         memcpy(cur,head,sizeof(cur));//当前弧优化
         ret+=dfs(s,inf);
     }
     return ret;
 }
 int main(){
     n=read(),m=read();
     rep (i,,m){
         int x=read(),y=read(),a=read(),b=read();
         ins(x,y,a,b-a);d[x]-=a,d[y]+=a;
     }
     s=n+,t=n+;
     rep (i,,n) if (d[i]>) ins(s,i,,d[i]);else ins(i,t,,-d[i]);
     dinic();
     for (int i=head[s];i;i=e[i].nxt) if (vis[e[i].to]!=-){puts("NO");exit();}
     puts("YES");
     rep (i,,m) printf("%d\n",e[i*+].d+e[i*+].c);
     return ;
 }