• FJ省队集训DAY3 T2


    思路:如果一个DAG要的路径上只要一条边去切掉,那么要怎么求?很容易就想到最小割,但是如果直接做最小割会走出重复的部分,那我们就这样:反向边设为inf,这样最小割的时候就不会割到了,判断无解我们直接用tarjan

      1 #include<cstdio>
      2 #include<iostream>
      3 #include<cmath>
      4 #include<cstring>
      5 #include<algorithm>
      6 #define ll long long
      7 const ll inf = 1ll << 60;
      8 struct edge{
      9     int u,v,w;
     10 }e[200005];
     11 ll flow[200005];
     12 int op[200005],dis[200005],cnt[200005],a[115][115],pd[10005];
     13 int tot,go[200005],next[200005],first[200005],S,T,nodes,sz;
     14 int n,m,vis[200005],instack[200005],c[200005],top,dfn[200005],low[200005],belong[200005],num;
     15 int read(){
     16     int t=0,f=1;char ch=getchar();
     17     while (ch<'0'||ch>'9'){if (ch=='-')f=-1;ch=getchar();}
     18     while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();}
     19     return t*f;
     20 }
     21 void insert(int x,int y){
     22     tot++;
     23     go[tot]=y;
     24     next[tot]=first[x];
     25     first[x]=tot;
     26 }
     27 void insert(int x,int y,ll z){
     28     tot++;
     29     go[tot]=y;
     30     next[tot]=first[x];
     31     first[x]=tot;
     32     flow[tot]=z;
     33 }
     34 void add(int x,int y,ll z){
     35     insert(x,y,z);op[tot]=tot-1;
     36     insert(y,x,inf);op[tot]=tot+1;
     37 }
     38 void tarjan(int x){
     39     vis[x]=instack[x]=1;
     40     c[++top]=x;dfn[x]=low[x]=++sz;
     41     for (int i=first[x];i;i=next[i]){
     42         int pur=go[i];
     43         if (!vis[pur]){
     44             tarjan(pur);
     45             low[x]=std::min(low[x],low[pur]);
     46         }else if (instack[pur]){
     47             low[x]=std::min(low[x],dfn[pur]);
     48         }
     49     }
     50     if (low[x]==dfn[x]){
     51         num++;
     52         while (c[top]!=x){
     53             belong[c[top]]=num;
     54             instack[c[top]]=0;
     55             top--;
     56         }
     57             belong[c[top]]=num;
     58             instack[c[top]]=0;
     59             top--;
     60     }
     61 }
     62 ll dfs(int x,ll f){
     63     if (x==T) return f;
     64     int mn=nodes;ll sum=0;
     65     for (int i=first[x];i;i=next[i]){
     66         int pur=go[i];
     67         if (flow[i]&&dis[pur]+1==dis[x]){
     68             ll F=std::min(f-sum,flow[i]);
     69             ll save=dfs(pur,F);
     70             flow[i]-=save;
     71             flow[op[i]]+=save;
     72             sum+=save;
     73             if (dis[S]>=nodes||f==sum) return sum;
     74         }
     75         if (flow[i]) mn=std::min(mn,dis[pur]);
     76     }
     77     if (sum==0){
     78         cnt[dis[x]]--;
     79         if (cnt[dis[x]]==0){
     80             dis[S]=nodes;
     81         }else{
     82             dis[x]=mn+1;
     83             cnt[dis[x]]++;
     84         }
     85     }
     86     return sum;
     87 }
     88 int main(){
     89     n=read();m=read();
     90     for (int i=1;i<=m;i++){
     91         e[i].u=read()+1,e[i].v=read()+1,e[i].w=read();
     92         a[e[i].u][e[i].v]=1;
     93         insert(e[i].u,e[i].v);
     94     }
     95     for (int i=1;i<=n;i++)
     96      if (!vis[i]) tarjan(i);
     97     if (belong[1]==belong[n]){
     98         puts("-1");
     99         return 0;
    100     } 
    101     for (int i=1;i<=n;i++)
    102      a[i][i]=1;
    103     for (int k=1;k<=n;k++)
    104      for (int i=1;i<=n;i++)
    105       for (int j=1;j<=n;j++) 
    106        a[i][j]|=a[i][k]&&a[k][j];
    107     for (int i=1;i<=n;i++)
    108      if (a[1][i]&&a[i][n]) pd[i]=1;
    109     for (int i=1;i<=m;i++)
    110      if (pd[e[i].u]&&pd[e[i].v])
    111       add(e[i].u,e[i].v,e[i].w);
    112     S=1;T=n;nodes=n;
    113     int ans=0;
    114     while (dis[S]<nodes&&ans<inf) ans+=dfs(S,inf);
    115     if (ans==0) printf("-1");
    116     else printf("%d
    ",ans);      
    117 }
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  • 原文地址:https://www.cnblogs.com/qzqzgfy/p/5645067.html
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