FJ省队集训DAY3 T2

Posted 2020-08-03 GFY

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\\n",ans);      
117 }