Farmer John's cows like to keep in touch via email so they have created a network of cowputers so that they can intercowmunicate. These machines route email so that if there exists a sequence of c cowputers a1, a2, ..., a(c) such that a1 is connected to a2, a2 is connected to a3, and so on then a1 and a(c) can send email to one another.
Unfortunately, a cow will occasionally step on a cowputer or Farmer John will drive over it, and the machine will stop working. This means that the cowputer can no longer route email, so connections to and from that cowputer are no longer usable.
Two cows are pondering the minimum number of these accidents that can occur before they can no longer use their two favorite cowputers to send email to each other. Write a program to calculate this minimal value for them, and to calculate a set of machines that corresponds to this minimum.
For example the network:
1
/
3 - 2
shows 3 cowputers connected with 2 lines. We want to send messages between cowputers 1 and 2. Direct lines connect 1-3 and 2-3. If cowputer 3 is down, them there is no way to get a message from 1 to 2.
PROGRAM NAME: telecow
INPUT FORMAT
| Line 1 | Four space-separated integers: N, M, c1, and c2. N is the number of computers (1 <= N <= 100), which are numbered 1..N. M is the number of connections between pairs of cowputers (1 <= M <= 600). The last two numbers, c1 and c2, are the id numbers of the cowputers that the communicating cows are using. Each connection is unique and bidirectional (if c1 is connected to c2, then c2 is connected to c1). There can be at most one wire between any two given cowputers. Computers c1 and c2 will not have a direct connection. |
| Lines 2..M+1 | The subsequent M lines contain pairs of cowputers id numbers that have connections between them. |
SAMPLE INPUT (file telecow.in)
3 2 1 2 1 3 2 3
OUTPUT FORMAT
Generate two lines of output. The first line is the minimum number of (well-chosen) cowputers that can be down before terminals c1 & c2 are no longer connected. The second line is a minimal-length sorted list of cowputers that will cause c1 & c2 to no longer be connected. Note that neither c1 nor c2 can go down. In case of ties, the program should output the set of computers that, if interpreted as a base N number, is the smallest one.
SAMPLE OUTPUT (file telecow.out)
1 3
——————————————————————————————题解
拆点求最小割，每个点相当于一个从u*2-1->u*2的边，为了保证字典序最小，我们把它的权值赋成600*101+u
一条无向边u,v，使u，v两条边首尾相连，u*2->v*2-1,v*2->u*2-1
从原点做floodfill求割
1 /* 2 ID: ivorysi 3 LANG: C++ 4 PROG: telecow 5 */ 6 #include <iostream> 7 #include <cstdio> 8 #include <cstring> 9 #include <algorithm> 10 #include <queue> 11 #include <set> 12 #include <vector> 13 #include <string.h> 14 #include <cmath> 15 #include <stack> 16 #include <map> 17 #define siji(i,x,y) for(int i=(x);i<=(y);++i) 18 #define gongzi(j,x,y) for(int j=(x);j>=(y);--j) 19 #define xiaosiji(i,x,y) for(int i=(x);i<(y);++i) 20 #define sigongzi(j,x,y) for(int j=(x);j>(y);--j) 21 #define inf 0x5f5f5f5f 22 #define ivorysi 23 #define mo 97797977 24 #define hash 974711 25 #define base 47 26 #define pss pair<string,string> 27 #define MAXN 5000 28 #define fi first 29 #define se second 30 #define pii pair<int,int> 31 #define esp 1e-8 32 typedef long long ll; 33 using namespace std; 34 int f[105][105]; 35 struct data { 36 int to,next,val; 37 }edge[10005]; 38 int head[205],sumedge=1; 39 void add(int u,int v,int c) { 40 edge[++sumedge].to=v; 41 edge[sumedge].next=head[u]; 42 edge[sumedge].val=c; 43 head[u]=sumedge; 44 } 45 void addtwo(int u,int v,int c) { 46 add(u,v,c); 47 add(v,u,0); 48 } 49 int n,m,c1,c2,vis[205],mincut; 50 vector<int> temp,ans,list; 51 void dfs(int u) { 52 if(vis[u]==1) return; 53 vis[u]=1; 54 for(int i=head[u];i;i=edge[i].next) { 55 if(edge[i].val!=0) dfs(edge[i].to); 56 } 57 } 58 int dis[205],gap[205]; 59 int sap(int u,int aug) { 60 if(u==c2*2) return aug; 61 int flow=0,dmin=2*n-1; 62 63 for(int i=head[u];i;i=edge[i].next) { 64 int v=edge[i].to; 65 if(edge[i].val>0) { 66 if(dis[u]==dis[v]+1) { 67 int t=sap(v,min(edge[i].val,aug-flow)); 68 //应该流走的是aug-flow，为什么每次都是网络流写的出问题qwq 69 edge[i].val-=t; 70 edge[i^1].val+=t; 71 flow+=t; 72 if(flow==aug) break; 73 if(dis[c1*2-1]>=2*n) return flow; 74 } 75 dmin=min(dmin,dis[v]); 76 } 77 } 78 if(!flow){ 79 --gap[dis[u]]; 80 if(gap[dis[u]]==0) dis[c1*2-1]=2*n; 81 dis[u]=dmin+1; 82 ++gap[dis[u]]; 83 } 84 return flow; 85 } 86 void init() { 87 scanf("%d%d%d%d",&n,&m,&c1,&c2); 88 int u,v; 89 siji(i,1,m) { 90 scanf("%d%d",&u,&v); 91 addtwo(u*2,v*2-1,inf+10);addtwo(v*2,u*2-1,inf+10); 92 } 93 siji(i,1,n) { 94 if(i==c1||i==c2) addtwo(i*2-1,i*2,inf+10); 95 else addtwo(i*2-1,i*2,600*101+i); 96 } 97 while(dis[c1*2-1]<2*n) sap(c1*2-1,inf); 98 } 99 void solve() { 100 init(); 101 dfs(c1*2-1); 102 while(1) { 103 siji(i,1,2*n) { 104 if(vis[i]) { 105 for(int j=head[i];j;j=edge[j].next) { 106 if(!vis[edge[j].to]) { 107 list.push_back(edge[j].to); 108 if(j%2==0) 109 temp.push_back(edge[j^1].val%101); 110 } 111 } 112 } 113 } 114 if(temp.size()<1) break; 115 if(ans.size()<1) {ans=temp;sort(ans.begin(),ans.end());} 116 else { 117 sort(temp.begin(),temp.end()); 118 if(ans.size()>temp.size()) ans=temp; 119 else if(ans.size()==temp.size()) { 120 xiaosiji(i,0,temp.size()) { 121 if(temp[i]<ans[i]) {ans=temp;break;} 122 } 123 } 124 } 125 temp.clear(); 126 127 xiaosiji(i,0,list.size()) { 128 dfs(list[i]); 129 } 130 list.clear(); 131 } 132 printf("%d ",ans.size()); 133 xiaosiji(i,0,ans.size()) { 134 printf("%d%c",ans[i]," "[i==ans.size()-1]); 135 } 136 } 137 int main(int argc, char const *argv[]) 138 { 139 #ifdef ivorysi 140 freopen("telecow.in","r",stdin); 141 freopen("telecow.out","w",stdout); 142 #else 143 freopen("f1.in","r",stdin); 144 #endif 145 solve(); 146 return 0; 147 }