Sgu213-Strong (Defense)

Sgu213-Strong (Defense)

Question:

Give you an undirected graph, and then a second, t represents the start point and the end point, then input n, m, s, t, and divide m pieces of undirected edges into L sets, so that after the edge of any set is removed, it cannot reach s from t (or s to t. I don't remember it, but it's similar .....), Then you need to find the largest L and then output each edge set.

Solution:

First, I thought about the network stream, but later I thought It was actually very simple.

Run the shortest path at the starting point of s, and a dis [t] indicates the shortest distance from s to t, then L = dis [t]

Group the edge ~ Dis [t], edges (Assume that dis [u] <= dis [v]) If dis [u] + 1 = dis [v] is satisfied, add this edge to the dis [v] collection, and place other unsatisfied edges as needed.

The correctness of the specific algorithm is not proved.

AC code:

#include 
 
  #include 
  
   #include 
   
    #include 
    
     #include #include 
     
      #include 
      
       #define MAX(a,b) ((a)>(b)?(a):(b))#define MIN(a,b) ((a)>(b)?(b):(a))using namespace std;int n,m,s,t;struct bian_{int num;int next;}bian[160010]={{0,0}};int g[160010][2]={{0}};int First[410]={0};int dis[410]={0};int hash[410]={0};int dui[1000010]={0};inline void Add(int p,int q,int k){bian[k].num=q;bian[k].next=First[p];First[p]=k;return;}void SPFA(){int duip=1;memset(dis,0x3f3f3f3f,sizeof(dis));dis[s]=0;dui[duip]=s;hash[s]=1;for(int i=1;i<=duip;i++){for(int p=First[dui[i]];p!=0;p=bian[p].next){if(dis[dui[i]]+1
       
         ans[410];for(int i=1;i<=m;i++){int p=dis[g[i][0]],q=dis[g[i][1]];if(p>q) swap(p,q);if(q==p+1){if(q>=dis[t])ans[dis[t]].push_back(i);else ans[q].push_back(i);}else ans[dis[t]].push_back(i);}printf("%d\n",dis[t]);for(int i=1;i<=dis[t];i++){printf("%d ",ans[i].size());for(int j=ans[i].size()-1;j>=0;j--)printf("%d%c",ans[i][j],j==0?'\n':' ');}return 0;}