挺裸的一道最小割。把每台电脑拆成一条容量为1的边，然后就跑最大流。从小到大枚举每台电脑，假如去掉后 最大流=之前最大流+1，那这台电脑就是answer之一了。
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#include<cstdio>
#include<vector>
#include<cstring>
#define rep(i,r) for(int i=0;i<r;i++)
#define clr(x,c) memset(x,c,sizeof(x))
#define Rep(i,l,r) for(int i=l;i<r;i++)
using namespace std;
const int maxn=100*2+5;
const int inf=1<<30;
int x[maxn][2];
struct Edge {
int from,to,cap,flow;
Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f) {}
};
struct ISAP {
int n,m,s,t;
int p[maxn];
int cur[maxn];
int num[maxn];
int d[maxn];
vector<int> g[maxn];
vector<Edge> edges;
void init(int n) {
this->n=n;
rep(i,n) g[i].clear();
edges.clear();
}
int addEdge(int from,int to,int cap) {
edges.push_back( (Edge) {from,to,cap,0} );
edges.push_back( (Edge) {to,from,0,0} );
m=edges.size();
g[from].push_back(m-2);
g[to].push_back(m-1);
return m-2;
}
int augment() {
int x=t,a=inf;
while(x!=s) {
Edge &e=edges[p[x]];
a=min(a,e.cap-e.flow);
x=edges[p[x]].from;
}
x=t;
while(x!=s) {
edges[p[x]].flow+=a;
edges[p[x]^1].flow-=a;
x=edges[p[x]].from;
}
return a;
}
int maxFlow(int s,int t) {
int flow=0;
this->s=s; this->t=t;
clr(d,0); clr(num,0); clr(cur,0); clr(p,0);
rep(i,edges.size()) edges[i].flow=0;
rep(i,n) num[d[i]]++;
int x=s;
while(d[s]<n) {
if(x==t) { flow+=augment(); x=s; }
int ok=0;
Rep(i,cur[x],g[x].size()) {
Edge &e=edges[g[x][i]];
if(e.cap>e.flow && d[x]==d[e.to]+1) {
ok=1;
p[e.to]=g[x][i];
cur[x]=i;
x=e.to;
break;
}
}
if(!ok) {
int m=n-1;
rep(i,g[x].size()) {
Edge &e=edges[g[x][i]];
if(e.cap>e.flow) m=min(m,d[e.to]);
}
if(--num[d[x]]==0) break;
num[d[x]=m+1]++;
cur[x]=0;
if(x!=s) x=edges[p[x]].from;
}
}
return flow;
}
} isap;
int main()
{
freopen("telecow.in","r",stdin);
freopen("telecow.out","w",stdout);
clr(x,-1);
int n,m,s,t,a,b;
scanf("%d%d%d%d",&n,&m,&s,&t);
--s; --t;
isap.init(2*n);
rep(i,n) {
x[i][0]=isap.addEdge(i,i+n,1);
x[i][1]=isap.addEdge(i+n,i,1);
}
rep(i,m) {
scanf("%d%d",&a,&b);
--a; --b;
isap.addEdge(a+n,b,m+1);
isap.addEdge(b+n,a,m+1);
}
int maxFlow=isap.maxFlow(s+n,t);
vector<int> ans; ans.clear();
rep(i,n) if(i!=s && i!=t) {
isap.edges[x[i][0]].cap=0;
isap.edges[x[i][1]].cap=0;
if(isap.maxFlow(s+n,t)==maxFlow-1) {
ans.push_back(i+1);
maxFlow--;
} else {
isap.edges[x[i][0]].cap=1;
isap.edges[x[i][1]].cap=1;
}
if(!maxFlow) break;
}
printf("%d\n",ans.size());
printf("%d",ans[0]);
Rep(i,1,ans.size()) printf(" %d",ans[i]);
printf("\n");
return 0;
}
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Telecowmunication
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 1 with 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 questioning 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. Computer c1 and c2 will not have a direction 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 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