Flow Problem
Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 8199 Accepted Submission(s): 3814
Problem Description
Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.
Input
The first line of input contains an integer T, denoting the number of test cases.
For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)
Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)
For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)
Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)
Output
For each test cases, you should output the maximum flow from source 1 to sink N.
Sample Input
2
3 2
1 2 1
2 3 1
3 3
1 2 1
2 3 1
1 3 1
3 2
1 2 1
2 3 1
3 3
1 2 1
2 3 1
1 3 1
Sample Output
Case 1: 1
Case 2: 2
题意：给一个有向图，求源点为1汇点为n的最大流
思路：裸题，ISAP。。原来之前自己写的模版有点疏漏了。。
#include <vector>
#include <cstdio>
#include <cstring>
#include <queue>
#define FOR(i,n) for(i=1;i<=(n);i++)
using namespace std;
const int INF = 1e9;
const int N = ; struct Edge{
int from,to,cap,flow;
}; struct ISAP{
int n,m,s,t;
int p[N],num[N];
vector<Edge> edges;
vector<int> G[N];
bool vis[N];
int d[N],cur[N];
void init(int _n,int _m)
{
n=_n; m=_m;
int i;
edges.clear();
FOR(i,n)
{
G[i].clear();
d[i]=INF;
}
}
void AddEdge(int from,int to,int cap)
{
edges.push_back((Edge){from,to,cap,});
edges.push_back((Edge){to,from,,});
m = edges.size();
G[from].push_back(m-);
G[to].push_back(m-);
}
bool BFS()
{
memset(vis,,sizeof(vis));
queue<int> Q;
Q.push(t);
d[t]=;
vis[t]=;
while(!Q.empty())
{
int x = Q.front(); Q.pop();
for(unsigned i=;i<G[x].size();i++)
{
Edge& e = edges[G[x][i]^];
if(!vis[e.from] && e.cap>e.flow)
{
vis[e.from]=;
d[e.from] = d[x]+;
Q.push(e.from);
}
}
}
return vis[s];
}
int Augment()
{
int x=t, a=INF;
while(x!=s)
{
Edge& e = edges[p[x]];
a = min(a,e.cap-e.flow);
x = edges[p[x]].from;
}
x = t;
while(x!=s)
{
edges[p[x]].flow+=a;
edges[p[x]^].flow-=a;
x=edges[p[x]].from;
}
return a;
}
int Maxflow(int _s,int _t)
{
s=_s; t=_t;
int flow = , i;
BFS();
if(d[s]>=n) return ;
memset(num,,sizeof(num));
memset(p,,sizeof(p));
FOR(i,n) if(d[i]<INF) num[d[i]]++;
int x=s;
memset(cur,,sizeof(cur));
while(d[s]<n)
{
if(x==t)
{
flow+=Augment();
x=s;
}
int ok=;
for(unsigned i=cur[x];i<G[x].size();i++)
{
Edge& e=edges[G[x][i]];
if(e.cap>e.flow && d[x]==d[e.to]+)
{
ok=;
p[e.to]=G[x][i];
cur[x]=i;
x=e.to;
break;
}
}
if(!ok)
{
int m=n-;
for(unsigned i=;i<G[x].size();i++)
{
Edge& e=edges[G[x][i]];
if(e.cap>e.flow) m=min(m,d[e.to]);
}
if(--num[d[x]]==) break;
num[d[x]=m+]++;
cur[x]=;
if(x!=s) x=edges[p[x]].from;
}
}
return flow;
}
}; ISAP isap; void run()
{
int n,m,u,v,c;
scanf("%d%d",&n,&m);
isap.init(n,m);
while(m--)
{
scanf("%d%d%d",&u,&v,&c);
isap.AddEdge(u,v,c);
//isap.AddEdge(v,u,c);
}
static int cas = ;
printf("Case %d: %d\n",cas++,isap.Maxflow(,n));
} int main()
{
freopen("case.txt","r",stdin);
int _;
scanf("%d",&_);
while(_--)
run();
return ;
}