题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=6214
思路:
1.要求在最小割的情况下割集边数最小,对于两种约束条件,可通过一定方法将其转化为单约束。
2.令边权值为w*MAX+1,MAX为一较大数,使得当割不同时,边权值作用最大;当割相同时,边数起作用。
3.求最大流,边数即为maxFlow%MAX。
#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn = 400+50;
const int INF = 0x3f3f3f3f;
struct Edge
{
int from, to;
LL cap, flow;
Edge(int a,int b,LL c,LL d):from(a),to(b),cap(c),flow(d) {}
};
struct Dinic
{
int n, m, s, t;
vector<Edge> edges;
vector<int> G[maxn];
bool vis[maxn];
LL d[maxn];
int cur[maxn];
void init(int n)
{
this->n=n;
for(int i = 0; i < n; i++) G[i].clear();
edges.clear();
}
void addEdge(int from, int to, LL 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);
}
bool BFS()
{
memset(vis, 0, sizeof(vis));
queue<int> Q;
Q.push(s);
vis[s] = 1;
d[s] = 0;
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, LL a)
{
if(x == t || a == 0) return a;
LL 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 == 0) break;
}
}
return flow;
}
LL MaxFlow(int s, int t)
{
this->s = s;
this->t = t;
LL flow = 0;
while(BFS())
{
memset(cur, 0, sizeof(cur));
flow += DFS(s, INF);
}
return flow;
}
};
Dinic g;
int n,m,s,t;
int main()
{
#ifdef debu
freopen("in.txt","r",stdin);
#endif // debug
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
scanf("%d%d",&s,&t);
g.init(n+1);
for(int i=0;i<m;i++)
{
int x,y,w;
scanf("%d%d%d",&x,&y,&w);
g.addEdge(x,y,(LL)w*1001+1);
}
printf("%lld\n",g.MaxFlow(s,t)%1001);
}
return 0;
}