题意
n个点,然后给出m条边,然后问删至少多少个点可以得到不连通的图。
思路
不连通的图,把一个分成两个连通图是删点最少的情况,然后就联想到网络流最小割问题,最小割是删边,这个是删点,所以可以把一个点一切两开中间放一条边。然后任选一个点为起点,枚举其他的作为汇点,跑n-1次Dinic取最小值就行了。
代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 200;
struct Edge {
int v, cap, nxt;
};
int n, m;
int head[maxn],tot;
Edge edges[maxn*maxn];
void init() {
tot=0;
memset(head, -1, sizeof(head));
}
void addEdge(int u, int v, int cap) {
edges[tot].v = v, edges[tot].cap = cap, edges[tot].nxt = head[u], head[u] = tot++;
edges[tot].v = u, edges[tot].cap = 0, edges[tot].nxt = head[v], head[v] = tot++;
}
int d[maxn];
bool bfs(int s, int t) {
memset(d, -1, sizeof(d));
queue<int>q;d[s]=0;
q.push(s);
while(!q.empty()) {
int u = q.front(); q.pop();
if(u == t) return true;
for(int e = head[u]; ~e; e = edges[e].nxt) {
int &v = edges[e].v, cap = edges[e].cap;
if(d[v] == -1 && cap > 0) {
d[v] = d[u] + 1;
q.push(v);
}
}
}
return 0;
}
int dfs(int s, int t, int flow) {
if(s == t) return flow;
int pre = 0;
for(int e = head[s]; ~e; e = edges[e].nxt) {
int & v = edges[e].v, cap = edges[e].cap;
if(d[v] == d[s] + 1 && cap > 0) {
int tmp = min(flow-pre, cap);
int tf = dfs(v, t, tmp);
edges[e].cap -= tf;
edges[e^1].cap += tf;
pre += tf;
if(pre == flow) return pre;
}
}
return pre;
}
int dinic(int s, int t) {
int ret = 0;
while(bfs(s, t)) ret += dfs(s,t,inf);
return ret;
}
vector<pair<int, int> > vecp;
Edge tmp[maxn*maxn];
int main()
{
// freopen("/Users/maoxiangsun/MyRepertory/acm/i.txt", "r", stdin);
while(~scanf("%d%d", &n, &m)) {
init();
for(int i = 1; i < n; i++) addEdge(i, i+n, 1);
for(int i = 1; i <= m; i++) {
int u, v;
scanf(" (%d,%d)", &u, &v);
addEdge(u+n, v, inf);
addEdge(v+n, u, inf);
}
int ans = n;
int S = n;
memcpy(tmp, edges, sizeof(tmp));
for(int T = 1; T < n; T++) {
memcpy(edges, tmp, sizeof(tmp));
ans = min(ans, dinic(S,T));
}
printf("%d\n", ans);
}
return 0;
}
/*
0 0
1 0
3 3 (0,1) (0,2) (1,2)
2 0
5 7 (0,1) (0,2) (1,3) (1,2) (1,4) (2,3) (3,4)
*/