多源汇最大流
很简单,建一个超级源点连接所有的源点,建一个超级汇点连接所有的汇点。
/*
* @Author: zhl
* @Date: 2020-10-20 11:09:59
*/
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i = a;i <= b;i++)
#define repE(i,u) for(int i = head[u];i;i = E[i].next)
//#define int long long
using namespace std;
const int N = 2e6 + 10, M = 2e6 + 10, inf = 1e9;
int n, m, s, t, tot, head[N];
int ans, dis[N], cur[N];
struct Edge {
int to, next, flow;
}E[M<<1];
void addEdge(int from, int to, int w) {
E[tot] = Edge{ to,head[from],w };
head[from] = tot++;
E[tot] = Edge{ from,head[to],0 };
head[to] = tot++;
}
int bfs() {
for (int i = 0; i <= n + 1; i++) dis[i] = -1;
queue<int>Q;
Q.push(s);
dis[s] = 0;
cur[s] = head[s];
while (!Q.empty()) {
int u = Q.front();
Q.pop();
for (int i = head[u]; ~i; i = E[i].next) {
int v = E[i].to;
if (E[i].flow && dis[v] == -1) {
Q.push(v);
dis[v] = dis[u] + 1;
cur[v] = head[v];
if (v == t)return 1; //分层成功
}
}
}
return 0;
}
int dfs(int x, int sum) {
if (x == t)return sum;
int k, res = 0;
for (int i = cur[x]; ~i && res < sum; i = E[i].next) {
cur[x] = i;
int v = E[i].to;
if (E[i].flow > 0 && (dis[v] == dis[x] + 1)) {
k = dfs(v, min(sum, E[i].flow));
if (k == 0) dis[v] = -1; //不可用
E[i].flow -= k;E[i ^ 1].flow += k;
res += k;sum -= k;
}
}
return res;
}
int Dinic() {
int ans = 0;
while (bfs()) {
ans += dfs(s,inf);
}
return ans;
}
signed main() {
scanf("%d%d%d%d",&n,&m,&s,&t);
memset(head, -1, sizeof(int) * (n + 10));
rep(i,1,s){
int x;scanf("%d",&x);
addEdge(0,x,inf);
}
rep(i,1,t){
int x;scanf("%d",&x);
addEdge(x,n+1,inf);
}
rep(i,1,m){
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
addEdge(a,b,c);
}
s = 0,t = n + 1;
printf("%d
",Dinic());
}
关键边
如果增加某一条边的容量可以增加最大流,则为关键边
跑一个最大流后,从源点bfs,从汇点bfs。交接处的边就是关键边
for (int i = 0; i < tot; i+=2) {
if (!E[i].flow and vis_t[E[i].to] and vis_s[E[i ^ 1].to]) ans++;
}
这里要 i+=2
/*
* @Author: zhl
* @Date: 2020-10-21 09:45:30
*/
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i = a;i <= b;i++)
#define repE(i,u) for(int i = head[u];~i;i = E[i].next)
using namespace std;
const int N = 5e2 + 10, M = 5e3 + 10, inf = 1e9;
struct Edge {
int to, flow, next;
}E[M << 1];
int head[N], tot;
void addEdge(int from, int to, int w) {
E[tot] = Edge{ to,w,head[from] };
head[from] = tot++;
E[tot] = Edge{ from,0,head[to] };
head[to] = tot++;
}
int n, m, s, t;
int dis[N], cur[N];
bool bfs() {
rep(i, 0, n)dis[i] = -1;
queue<int>Q;
Q.push(s); dis[s] = 0; cur[s] = head[s];
while (!Q.empty()) {
int u = Q.front(); Q.pop();
repE(i, u) {
int v = E[i].to;
if (dis[v] == -1 and E[i].flow) {
cur[v] = head[v];
dis[v] = dis[u] + 1;
Q.push(v);
if (v == t)return true;
}
}
}
return false;
}
int dfs(int u, int limit) {
if (u == t)return limit;
int k, res = 0;
for (int i = cur[u]; ~i and res < limit; i = E[i].next) {
int v = E[i].to;
cur[u] = i;
if (dis[v] == dis[u] + 1 and E[i].flow) {
k = dfs(v, min(limit, E[i].flow));
if (k == 0)dis[v] = -1;
E[i].flow -= k; E[i ^ 1].flow += k;
limit -= k; res += k;
}
}
return res;
}
int Dinic() {
int res = 0;
while (bfs())res += dfs(s, inf);
return res;
}
int vis_s[N], vis_t[N];
void Dfs(int u, int* vis, int p) {
vis[u] = 1;
repE(i, u) {
int v = E[i].to;
if (!vis[v] and E[i ^ p].flow) {
Dfs(v, vis, p);
}
}
}
int main() {
scanf("%d%d", &n, &m);
memset(head, -1, sizeof head);
rep(i, 1, m) {
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
addEdge(a + 1, b + 1, c);
}
s = 1, t = n;
Dinic();
Dfs(s, vis_s, 0); Dfs(t, vis_t, 1);
int ans = 0;
for (int i = 0; i < tot; i+=2) {
if (!E[i].flow and vis_t[E[i].to] and vis_s[E[i ^ 1].to]) {
ans++;
}
}
printf("%d
", ans);
}