• POJ3184 Ikki's Story I


    求一次最大流后,分别对所有满流的边的容量+1,然后看是否存在增广路。

      1 #include<cstdio>
      2 #include<cstring>
      3 #include<queue>
      4 #include<algorithm>
      5 using namespace std;
      6 #define INF (1<<30)
      7 #define MAXN 555 
      8 #define MAXM 11111 
      9 
     10 struct Edge{
     11     int v,cap,flow,next;
     12 }edge[MAXM];
     13 int vs,vt,NE,NV;
     14 int head[MAXN];
     15 
     16 void addEdge(int u,int v,int cap){
     17     edge[NE].v=v; edge[NE].cap=cap; edge[NE].flow=0;
     18     edge[NE].next=head[u]; head[u]=NE++;
     19     edge[NE].v=u; edge[NE].cap=0; edge[NE].flow=0;
     20     edge[NE].next=head[v]; head[v]=NE++;
     21 }
     22 
     23 int level[MAXN];
     24 int gap[MAXN];
     25 void bfs(){
     26     memset(level,-1,sizeof(level));
     27     memset(gap,0,sizeof(gap));
     28     level[vt]=0;
     29     gap[level[vt]]++;
     30     queue<int> que;
     31     que.push(vt);
     32     while(!que.empty()){
     33         int u=que.front(); que.pop();
     34         for(int i=head[u]; i!=-1; i=edge[i].next){
     35             int v=edge[i].v;
     36             if(level[v]!=-1) continue;
     37             level[v]=level[u]+1;
     38             gap[level[v]]++;
     39             que.push(v);
     40         }
     41     }
     42 }
     43 
     44 int pre[MAXN];
     45 int cur[MAXN];
     46 int ISAP(bool statue){
     47     bfs();
     48     memset(pre,-1,sizeof(pre));
     49     memcpy(cur,head,sizeof(head));
     50     int u=pre[vs]=vs,flow=0,aug=INF;
     51     gap[0]=NV;
     52     while(level[vs]<NV){
     53         bool flag=false;
     54         for(int &i=cur[u]; i!=-1; i=edge[i].next){
     55             int v=edge[i].v;
     56             if(edge[i].cap!=edge[i].flow && level[u]==level[v]+1){
     57                 flag=true;
     58                 pre[v]=u;
     59                 u=v;
     60                 //aug=(aug==-1?edge[i].cap:min(aug,edge[i].cap));
     61                 aug=min(aug,edge[i].cap-edge[i].flow);
     62                 if(v==vt){
     63                     if(statue&&aug) return 1;
     64                     flow+=aug;
     65                     for(u=pre[v]; v!=vs; v=u,u=pre[u]){
     66                         edge[cur[u]].flow+=aug;
     67                         edge[cur[u]^1].flow-=aug;
     68                     }
     69                     //aug=-1;
     70                     aug=INF;
     71                 }
     72                 break;
     73             }
     74         }
     75         if(flag) continue;
     76         int minlevel=NV;
     77         for(int i=head[u]; i!=-1; i=edge[i].next){
     78             int v=edge[i].v;
     79             if(edge[i].cap!=edge[i].flow && level[v]<minlevel){
     80                 minlevel=level[v];
     81                 cur[u]=i;
     82             }
     83         }
     84         if(--gap[level[u]]==0) break;
     85         level[u]=minlevel+1;
     86         gap[level[u]]++;
     87         u=pre[u];
     88     }
     89     return flow;
     90 }
     91 int main(){
     92     int n,m,a,b,c;
     93     scanf("%d%d",&n,&m);
     94     vs=0; vt=n-1; NV=n; NE=0;
     95     memset(head,-1,sizeof(head));
     96     while(m--){
     97         scanf("%d%d%d",&a,&b,&c);
     98         addEdge(a,b,c);
     99     }
    100     ISAP(0);
    101     int cnt=0;
    102     for(int i=0; i<NE; i+=2){
    103         if(edge[i].flow!=edge[i].cap) continue;
    104         ++edge[i].cap;
    105         if(ISAP(1)) ++cnt;
    106         --edge[i].cap;
    107     }
    108     printf("%d",cnt);
    109     return 0;
    110 }
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  • 原文地址:https://www.cnblogs.com/WABoss/p/4851965.html
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