2017 ACM/ICPC Asia Regional Qingdao Online 1009(最大流Dinic算法)
2017-09-20 21:41
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Smallest Minimum Cut
题意:求最小割边数。
给权值hash一下即可。使用最大流的Dinic算法,时间复杂度O(m*n^2)。
代码:
#include <bits/stdc++.h>
using namespace std;
const int max_v=202;
const long long INF=1ll << 60;
const int MOD=100000;
int n,m,s,t,u,v;
long long w;
long long min(long long a,long long b)
{
if (a < b)
return a;
else
return b;
}
//用于表示边的结构体(终点,容量,反向边)
struct edge
{
int to;
long long cap;
int rev;
};
vector<edge> G[max_v];//图的邻接表表示
int level[max_v];//顶点到源点的距离标号
//向图中增加一条从s到t容量为cap的边
void add_edge(int from,int to,long long cap)
{
G[from].push_back((edge){to,cap,G[to].size()});
G[to].push_back((edge){from,0,G[from].size()-1});
}
bool bfs(int s, int t)
{
memset(level,-1,sizeof(level));
queue<int> que;
level[s]=0;
que.push(s);
while(!que.empty())
{
int v=que.front(); que.pop();
if(v==t)
{
return true;
}
for(int i=0;i<G[v].size();i++)
{
edge &e=G[v][i];
if (e.cap>0&&level[e.to]<0)
{
level[e.to]=level[v]+1;
que.push(e.to);
}
}
}
return false;
}
//通过DFS寻找增广路
long long dfs(int v,int t,long long f)
{
if(v==t) return f;
for(int i=0;i<G[v].size();i++)
{
edge &e=G[v][i];
if(e.cap>0&&level[v]+1==level[e.to])
{
long long d=dfs(e.to,t,min(f,e.cap));
if(d>0)
{
e.cap-=d;
G[e.to][e.rev].cap+=d;
return d;
}
}
}
return 0;
}
//求解从s到t的最大流
long long max_flow(int s,int t)
{
long long flow=0;
while(bfs(s,t))
{
flow+=dfs(s, t, INF);
}
return flow;
}
int main()
{
int cas;
scanf("%d", &cas);
while (cas--)
{
for(int i=0;i<max_v;i++)
G[i].clear();
scanf("%d%d", &n, &m);
scanf("%d%d", &s, &t);
for (int i = 1; i <= m; i++)
{
scanf("%d%d%lld", &u, &v, &w);
add_edge(u, v, w * MOD + 1);
}
printf("%lld\n", max_flow(s, t) % MOD);
}
return 0;
}
题意:求最小割边数。
给权值hash一下即可。使用最大流的Dinic算法,时间复杂度O(m*n^2)。
代码:
#include <bits/stdc++.h>
using namespace std;
const int max_v=202;
const long long INF=1ll << 60;
const int MOD=100000;
int n,m,s,t,u,v;
long long w;
long long min(long long a,long long b)
{
if (a < b)
return a;
else
return b;
}
//用于表示边的结构体(终点,容量,反向边)
struct edge
{
int to;
long long cap;
int rev;
};
vector<edge> G[max_v];//图的邻接表表示
int level[max_v];//顶点到源点的距离标号
//向图中增加一条从s到t容量为cap的边
void add_edge(int from,int to,long long cap)
{
G[from].push_back((edge){to,cap,G[to].size()});
G[to].push_back((edge){from,0,G[from].size()-1});
}
bool bfs(int s, int t)
{
memset(level,-1,sizeof(level));
queue<int> que;
level[s]=0;
que.push(s);
while(!que.empty())
{
int v=que.front(); que.pop();
if(v==t)
{
return true;
}
for(int i=0;i<G[v].size();i++)
{
edge &e=G[v][i];
if (e.cap>0&&level[e.to]<0)
{
level[e.to]=level[v]+1;
que.push(e.to);
}
}
}
return false;
}
//通过DFS寻找增广路
long long dfs(int v,int t,long long f)
{
if(v==t) return f;
for(int i=0;i<G[v].size();i++)
{
edge &e=G[v][i];
if(e.cap>0&&level[v]+1==level[e.to])
{
long long d=dfs(e.to,t,min(f,e.cap));
if(d>0)
{
e.cap-=d;
G[e.to][e.rev].cap+=d;
return d;
}
}
}
return 0;
}
//求解从s到t的最大流
long long max_flow(int s,int t)
{
long long flow=0;
while(bfs(s,t))
{
flow+=dfs(s, t, INF);
}
return flow;
}
int main()
{
int cas;
scanf("%d", &cas);
while (cas--)
{
for(int i=0;i<max_v;i++)
G[i].clear();
scanf("%d%d", &n, &m);
scanf("%d%d", &s, &t);
for (int i = 1; i <= m; i++)
{
scanf("%d%d%lld", &u, &v, &w);
add_edge(u, v, w * MOD + 1);
}
printf("%lld\n", max_flow(s, t) % MOD);
}
return 0;
}
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