码迷,mamicode.com
首页 > 其他好文 > 详细

ZOJ::Reactor Cooling

时间:2018-02-06 20:32:40      阅读:136      评论:0      收藏:0      [点我收藏+]

标签:int   sizeof   problem   res   main   附加   inline   http   ++   

题目链接

ZOJ::Reactor Cooling

题解

题目大意:
给定下界,找出一种可行流方案
跑那么无上下界可行流
每条边在可行流中的流量=容量下界+附加流中它的流量

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using std::queue;
inline int read() {
    int x=0;
    char c=getchar();
    while(c<'0'||c>'9') c=getchar();
    while(c<='9'&&c>='0')x=x*10+c-'0',c=getchar();
    return x;
}
const int maxn = 207;
const int maxm = 100007;
int head[maxn];
int minflow[maxm],flow[maxn];
struct node{
    int next,num,v,flow;
}edge[maxm];
int n,m,num=1;
void add_edge(int u,int v,int flow,int N) {
    edge[++num].v=v;edge[num].flow=flow,edge[num].next=head[u];head[u]=num,edge[num].num=N;
    edge[++num].v=u,edge[num].flow=0;edge[num].next=head[v],head[v]=num,edge[num].num=N;
}
int S=0,T;
int dis[maxn],q[maxn];
int lev[maxn],cur[maxn];
bool bfs() {
    memset(lev,-1,sizeof lev);
    memcpy(cur,head,sizeof head);
    queue<int>que;
    lev[S]=0;
    que.push(S);
    while(!que.empty()) {
        int u=que.front();que.pop();
        for(int i=head[u];i;i=edge[i].next) {
            int v=edge[i].v;
            if(edge[i].flow>0&&lev[v]<0) {
                lev[v]=lev[u]+1;
                que.push(v);
            }
        }
    }
    if(lev[T]!=-1)return true;
    else return false;
}
int dfs(int now,int flow) {
    if(now==T)return flow;
    int rest=0,delta;
    for(int &i=cur[now];i;i=edge[i].next) {
        int v=edge[i].v;
        if(lev[v]==lev[now]+1&&edge[i].flow>0) {
            delta=dfs(v,std::min(flow-rest,edge[i].flow));
            if(delta) {
                edge[i].flow-=delta;
                edge[i^1].flow+=delta;
                rest+=delta;if(rest==flow)break;
            }
        }
    }
    if(rest==flow)lev[now]=-1;
    return rest;
}
int Dinic() {
    int ret=0,tmp;
    while(bfs()) {
        ret+=dfs(S,0x3f3f3f3f);
    }
    return ret;
}
int ans[maxm];
void work() {
    n=read(),m=read();T=n+1;
    for(int a,b,c,i=1;i<=m;++i) {
        a=read(),b=read(),minflow[i]=read(),c=read();
        add_edge(a,b,c-minflow[i],i);
        flow[a]-=minflow[i],flow[b]+=minflow[i];
    }
    int sum=0;
    for(int i=1;i<=n;++i) {
        if(flow[i]<0) add_edge(i,T,-flow[i],0);
        else sum+=flow[i],add_edge(S,i,flow[i],0);
    }
    if(Dinic()==sum) {
        puts("YES");
        for(int i=1;i<=n;++i) {
            for(int j=head[i];j;j=edge[j].next) {
                if(j%2==0)continue;
                ans[edge[j].num]=edge[j].flow+minflow[edge[j].num];
            }       
        }
        for(int i=1;i<=m;++i)printf("%d\n",ans[i]);
    }
    else puts("NO");
}
int main() {
    int Q=read();
    for(;Q--;) {
        memset(head,0,sizeof head);
        memset(flow,0,sizeof(flow));
        work();
    }
    return 0;
}
    

ZOJ::Reactor Cooling

标签:int   sizeof   problem   res   main   附加   inline   http   ++   

原文地址:https://www.cnblogs.com/sssy/p/8423418.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!