P4126 [AHOI2009]最小割
题意
- 给一个有向图。两问
- 某条边是不是一定是最小割边
- 某条边是不是有可能是最小割边
思路
跑完最大流的残量图中
-
如果某条正向边<u, v>的剩余流量为0,并且uv不在一个强连通分量中,那么这个边可能是最小割边。
-
在可能是最小割的条件下,如果u和源点s在同一个强连通分量中,且v和汇点t在同一个强连通分量中,那么这个边一定是最小割边。【言外之意就是,每一个最小割都必包含这条边】
关于最小割
- 最小割=最大流
如何求最小割?
在残量网络中,从源点s出发,能够到达的点构成点集S{}, 其余点构成点集T{}. 连接两个点集的正向满流边就是最小割边。
CODE
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
inline int read()
{
int x = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') { if(c == '-') f = -f; c = getchar(); }
while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
return x * f;
}
const int maxN = 4003;
const int maxM = 60004;
int n, m, s, t;
struct EDGE{
int adj, from, to, w;
EDGE(int a = -1, int b = 0, int c = 0, int d = 0): adj(a), from(b), to(c), w(d) {}
}edge[maxM << 1];
int head[maxN], cur[maxN], cnt;
void add_edge(int u, int v, int w)
{
edge[cnt] = EDGE(head[u], u, v, w);
head[u] = cnt ++ ;
}
int deep[maxN];
bool bfs()
{
for(int i = 0; i <= n; ++ i ) deep[i] = 0, cur[i] = head[i];
queue<int>q;
q.push(s); deep[s] = 1;
while(!q.empty())
{
int u = q.front(); q.pop();
for(int i = head[u]; ~i; i = edge[i].adj)
{
int v = edge[i].to;
if(!deep[v] && edge[i].w)
{
deep[v] = deep[u] + 1;
q.push(v);
if(v == t) return true;
}
}
}
return false;
}
ll dfs(int u, int flow)
{
if(u == t || !flow) return flow;
for(int& i = cur[u]; ~i; i = edge[i].adj)
{
int v = edge[i].to;
if(deep[u] + 1 == deep[v] && edge[i].w)
{
if(ll nowFlow = dfs(v, min(flow, edge[i].w)))
{
edge[i].w -= nowFlow;
edge[i ^ 1].w += nowFlow;
return nowFlow;
}
}
}
return 0;
}
void dinic()
{
while(bfs()) while(dfs(s, INF));
}
int dfn[maxN], low[maxN], tot;
int belong[maxN], cnnct;
int Stack[maxN], top;
bool InSta[maxN];
void tarjan(int u)
{
dfn[u] = low[u] = ++ tot;
Stack[++ top] = u; InSta[u] = true;
for(int i = head[u]; ~i; i = edge[i].adj)
{
int v = edge[i].to;
if(edge[i].w == 0) continue;
if(!dfn[v])
{
tarjan(v);
low[u] = min(low[u], low[v]);
}
else if(InSta[v])
{
low[u] = min(low[u], dfn[v]);
}
}
if(low[u] == dfn[u])
{
++cnnct;
int now;
do{
now = Stack[top -- ];
InSta[now] = false;
belong[now] = cnnct;
}while(now != u);
}
}
int main()
{
memset(head, -1, sizeof(head));
n = read(); m = read(); s = read(); t = read();
for(int i = 0; i < m; ++ i )
{
int u, v, w;
u = read(); v = read(); w = read();
add_edge(u, v, w);
add_edge(v, u, 0);
}
dinic();
for(int i = 1; i <= n; ++ i ) if(!dfn[i]) tarjan(i);
for(int i = 0; i < cnt; i += 2)
{
int u = edge[i].from, v = edge[i].to;
if(belong[u] != belong[v] && edge[i].w == 0)
{
putchar('1');
putchar(' ');
if(belong[u] == belong[s] && belong[v] == belong[t])
putchar('1');
else putchar('0');
}
else printf("0 0");
putchar('\n');
}
return 0;
}