poj 1273 Drainage Ditches(最大费用流+最短增广路径算法)

Drainage Ditches

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 45239		Accepted: 17002

Description

Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's
clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.

Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.

Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.

Input

The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points
for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow
through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.
Output

For each case, output a single integer, the maximum rate at which water may emptied from the pond.
Sample Input

5 4
1 2 40
1 4 20
2 4 20
2 3 30
3 4 10

Sample Output

50

Source

USACO 93

思路：网络流入门，建图无难度。我用的是最短增广路径算法。复杂度是O（n*m*m）

#include<iostream>
#include<cstring>
using namespace std;
const int mm=210;
const int oo=1e9;
class node
{
public:int c,f;
node(){c=f=0;}
}e[mm][mm];
int pre[mm],le[mm][mm],q[mm];
int m,n,s,t,a,b,c;
void path_add()///找增广路径
{ memset(le,0,sizeof(le));
memset(pre,-1,sizeof(pre));
pre[1]=-2;int qs=0,qt=1;q[qs]=1;
int z;
while(qs!=qt&&pre[t]==-1)
{
z=q[qs++];qs%=mm;
for(int i=1;i<=n;i++)
{
if(pre[i]==-1)
{
if(e[z][i].c-e[z][i].f>0)///le[][]为残留网络
le[z][i]=e[z][i].c-e[z][i].f,pre[i]=z,q[qt++]=i,qt%=mm;
else if(e[i][z].f>0)
le[z][i]=e[i][z].f,pre[i]=z,q[qt++]=i,qt%=mm;
}
}
}
}
int flow_go()
{ if(pre[t]==-1)return 0;
int ret=oo,u,v;
for(u=pre[t],v=t;u!=-2;v=u,u=pre[u])///找出最小流
{
if(ret>le[u][v])ret=le[u][v];
}
if(ret>0)
{ u=pre[t];v=t;
while(u!=-2)///更改网络
{
if(e[u][v].c-e[u][v].f>0)
e[u][v].f+=ret;
else if(e[v][u].f>0)
e[u][v].f+=ret;
v=u;u=pre[u];
}
return ret;
}
else return 0;
}
int flow_max()
{
int ret=0,z;
while(1)
{
path_add();
z=flow_go();
ret+=z;
if(z==0)return ret;
}
}
int main()
{
while(cin>>m>>n)
{ memset(e,0,sizeof(e));
for(int i=0;i<m;i++)
{
cin>>a>>b>>c;
e[a][b].c+=c;
}s=1;t=n;
cout<<flow_max()<<"\n";
}
}