UVa 1660 Cable TV Network (最大流,最小割)
2024-09-03 18:03:29
题意:求一个无向图的点连通度。
析:把每个点拆成两个,然后中间连接一个容量为1的边,然后固定一个源点,枚举每个汇点,最小割。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std; typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-5;
const int maxn = 100 + 10;
const int mod = 1e6;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
struct Edge{
int from, to, cap, flow;
}; struct Dinic{
int m, s, t;
vector<Edge> edges;
vector<Edge> tmp;
vector<int> G[maxn];
bool vis[maxn];
int d[maxn];
int cur[maxn]; void init(){
edges.clear();
for(int i = 0; i < maxn; ++i) G[i].clear();
}
bool bfs(){
memset(vis, 0, sizeof vis);
queue<int> q;
q.push(s);
d[s] = 0; vis[s] = true;
while(!q.empty()){
int x = q.front(); q.pop();
for(int i = 0; i < G[x].size(); ++i){
Edge &e = edges[G[x][i]];
if(!vis[e.to] && e.cap > e.flow){
vis[e.to] = 1;
d[e.to] = d[x] + 1;
q.push(e.to);
}
}
}
return vis[t];
} int dfs(int x, int a){
if(x == t || a == 0) return a;
int flow = 0, f;
for(int& i = cur[x]; i < G[x].size(); ++i){
Edge& e = edges[G[x][i]];
if(d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0){
e.flow += f;
edges[G[x][i]^1].flow -= f;
flow += f;
a -= f;
if(!a) break;
}
}
return flow;
} int maxflow(int s, int t){
this->s = s; this->t = t;
int flow = 0;
while(bfs()){
memset(cur, 0, sizeof cur);
flow += dfs(s, INF);
}
return flow;
} void 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);
}
};
Dinic dinic; int main(){
while(scanf("%d %d", &n, &m) == 2){
dinic.init();
for(int i = 0; i < n; ++i) dinic.addEdge(i, i+n, 1);
for(int i = 0; i < m; ++i){
int u, v;
scanf(" (%d,%d)", &u, &v);
dinic.addEdge(u+n, v, INF);
dinic.addEdge(v+n, u, INF);
}
dinic.tmp = dinic.edges;
int ans = INF;
for(int i = 1; i < n; ++i){
ans = min(ans, dinic.maxflow(n, i));
dinic.edges = dinic.tmp;
}
printf("%d\n", ans == INF ? n : ans);
}
return 0;
}
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