LOJ #115. Passive sink has upstream and downstream feasible streams,

LOJ #115. Passive sink has upstream and downstream feasible streams,
#115. Description of a passive sink with upstream and downstream feasible streams

This is a template question.

N points, m edge, each side e has a lower flow threshold lower (e) \ text {lower} (e) lower (e) and the upper limit of traffic upper (e) \ text {upper} (e) upper (e), find a feasible solution so that all points meet the traffic balance conditions, all edges meet the traffic limit.

Input Format

The first line has two positive integers n and m.

Next m rows, each row has four integers: s, t, lower \ text {lower} lower, and upper \ text {upper} upper.

Output Format

If there is no solution, output a lineNO.

Otherwise, the first line is output.YESAnd then an integer in each line of m, indicating the traffic of each edge.

Sample input 1

4 61 2 1 22 3 1 23 4 1 24 1 1 21 3 1 24 2 1 2

Sample output 1

NO

Sample input 2

4 61 2 1 32 3 1 33 4 1 34 1 1 31 3 1 34 2 1 3

Sample output 2

YES123211

Data range and prompt

1 ≤ n ≤ 10200, 1 ≤ m ≤ 200 1 \ leq n \ leq 10200, 1 \ leq m \ leq 10200 1 ≤ n ≤, 1 ≤ m ≤

Show category labels

 

The question is not detailed. You can sort it out when you are free.

 

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<queue>using namespace std;const int MAXN=2000001;inline char nc(){    static char buf[MAXN],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,MAXN,stdin),p1==p2)?EOF:*p1++;}inline int read(){    char c=nc();int x=0,f=1;    while(c<'0'||c>'9'){if(c=='-')f=-1;c=nc();}    while(c>='0'&&c<='9'){x=x*10+c-'0';c=nc();}    return x*f;}int n,m,s,t;struct node{    int u,v,flow,nxt;}edge[MAXN];int head[MAXN],cur[MAXN],A[MAXN];int num=0;void AddEdge(int x,int y,int z){    edge[num].u=x;    edge[num].v=y;    edge[num].flow=z;    edge[num].nxt=head[x];    head[x]=num++;}void add_edge(int x,int y,int z){    AddEdge(x,y,z);    AddEdge(y,x,0);}int deep[MAXN],L[MAXN];bool BFS(){    memset(deep,0,sizeof(deep));    deep[s]=1;    queue<int>q;    q.push(s);    while(q.size()!=0)    {        int p=q.front();        q.pop();        for(int i=head[p];i!=-1;i=edge[i].nxt)            if(!deep[edge[i].v]&&edge[i].flow)            {                deep[edge[i].v]=deep[edge[i].u]+1;q.push(edge[i].v);                if(edge[i].v==t) return 1;            }                    }    return deep[t];    }int DFS(int now,int nowflow){    if(now==t||nowflow<=0)        return nowflow;    int totflow=0;    for(int &i=cur[now];i!=-1;i=edge[i].nxt)    {        if(deep[edge[i].v]==deep[edge[i].u]+1&&edge[i].flow)        {            int canflow=DFS(edge[i].v,min(nowflow,edge[i].flow));            edge[i].flow-=canflow;            edge[i^1].flow+=canflow;            totflow+=canflow;            nowflow-=canflow;            if(nowflow<=0)                break;        }    }    return totflow;}int Dinic(){    int ans=0;    while(BFS())    {        for(int i=0;i<=n;i++)            cur[i]=head[i];        ans+=DFS(s,1e8);    }    return ans;}int main(){    #ifdef WIN32    freopen("a.in","r",stdin);    #else    #endif    n=read();m=read();s=0;t=n+1;    memset(head,-1,sizeof(head));    for(int i=1;i<=m;i++)    {        int x=read(),y=read(),lower=read(),upper=read();L[i-1]=lower;        add_edge(x,y,upper-lower);A[x]-=lower;A[y]+=lower;    }    int sum=0;    for(int i=1;i<=n;i++)    {        if(A[i]>0) sum+=A[i],add_edge(s,i,A[i]);        else add_edge(i,t,-A[i]);    }    if(Dinic()!=sum) printf("NO");    else    {        printf("YES\n");        for(int i=0;i<m;i++)            printf("%d\n",edge[i*2|1].flow+L[i]);    }    return  0;}