Drainage Ditches
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8599 Accepted Submission(s): 4005
Problem DescriptionEvery 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.
InputThe 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.
OutputFor 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
SourceUSACO 93 解题报告EK模板，不多说。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#define inf 99999999
#define N 220
#define M 220
using namespace std;
int n,m,edge[N][N],flow,a[N],p[N];
queue<int>Q;
void ek()
{
while(1)
{
while(!Q.empty())
Q.pop();
memset(a,0,sizeof(a));
memset(p,0,sizeof(p));
Q.push(1);
a[1]=inf;
p[1]=1;
while(!Q.empty())
{
int u=Q.front();
Q.pop();
for(int v=1;v<=n;v++)
{
if(!a[v]&&edge[u][v]>0)
{
a[v]=min(a[u],edge[u][v]);
p[v]=u;
Q.push(v);
}
}
if(a[n])break;
}
if(!a[n])break;
for(int u=n;u!=1;u=p[u])
{
edge[p[u]][u]-=a[n];
edge[u][p[u]]+=a[n];
}
flow+=a[n];
}
}
int main()
{
int i,j,u,v,w;
while(~scanf("%d%d",&m,&n))
{
flow=0;
memset(edge,0,sizeof(edge));
for(i=0;i<m;i++)
{
scanf("%d%d%d",&u,&v,&w);
edge[u][v]+=w;
}
ek();
printf("%d\n",flow);
}
}
又学了Dinic算法。
。。
（基于邻接矩阵的）。前向星不知道怎么处理反向弧。。。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#define inf 99999999
#define N 200
#define M 200
using namespace std;
int edge[N][N],flow,l[N],n,m;
int bfs()
{
memset(l,-1,sizeof(l));
queue<int>Q;
l[1]=0;
Q.push(1);
while(!Q.empty())
{
int u=Q.front();
Q.pop();
for(int v=1;v<=n;v++)
{
if(edge[u][v]&&l[v]==-1)
{
l[v]=l[u]+1;
Q.push(v);
}
}
}
if(l[n]>0)
return 1;
else return 0;
}
int dfs(int x,int f)
{
if(x==n)return f;
int a;
for(int i=1;i<=n;i++)
{
if(edge[x][i]&&(l[i]==l[x]+1)&&(a=dfs(i,min(f,edge[x][i]))))
{
edge[x][i]-=a;
edge[i][x]+=a;
return a;
}
}
return 0;
}
int main()
{
int i,j,u,v,w;
while(~scanf("%d%d",&m,&n))
{
memset(edge,0,sizeof(edge));
for(i=0;i<m;i++)
{
scanf("%d%d%d",&u,&v,&w);
edge[u][v]+=w;
}
int a=0;
flow=0;
while(bfs())
while(a=dfs(1,inf))
flow+=a;
printf("%d\n",flow);
}
return 0;
}
最终觉醒了，之前看了一串非递归的前向星dinic，，。老认为不正确。，，换了一种思路，写成递归调用dfs和邻接表一样，对于处理反向弧加流的话在建边的时候处理，建边一次建两条，正向和反向，这样就知道。反向弧在哪里了，，。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#define N 300
#define M 30000
#define inf 99999999
using namespace std;
struct node
{
int u,v,w,r,next;
}edge[M];int n,m,head[N],l[N],cnt;
void add(int u,int v,int w)
{
edge[cnt].u=u;
edge[cnt].v=v;
edge[cnt].w=w;
edge[cnt].next=head[u];
edge[cnt].r=cnt+1;
head[u]=cnt++;
edge[cnt].u=v;
edge[cnt].v=u;
edge[cnt].w=0;
edge[cnt].next=head[v];
edge[cnt].r=cnt-1;
head[v]=cnt++;
}
int bfs()
{
queue<int >Q;
while(!Q.empty())
Q.pop();
Q.push(1);
memset(l,-1,sizeof(l));
l[1]=0;
while(!Q.empty())
{
int u=Q.front();
Q.pop();
for(int i=head[u];i!=-1;i=edge[i].next)
{
if(edge[i].w&&l[edge[i].v]==-1)
{
l[edge[i].v]=l[u]+1;
Q.push(edge[i].v);
}
}
}
if(l[n]>0)return 1;
else return 0;
}
int dfs(int x,int f)
{
int i,a;
if(x==n)return f;
for(i=head[x];i!=-1;i=edge[i].next)
{
if(edge[i].w&&l[edge[i].v]==l[x]+1&&(a=dfs(edge[i].v,min(f,edge[i].w))))
{
edge[i].w-=a;
edge[edge[i].r].w+=a;
return a;
}
}
return 0;
}
int main()
{
int i,j,u,v,w;
while(cin>>m>>n)
{
memset(head,-1,sizeof(head));
memset(edge,0,sizeof(edge));
cnt=0;
for(i=0;i<m;i++)
{
cin>>u>>v>>w;
add(u,v,w);
}
int ans=0,a;
while(bfs())
while(a=dfs(1,inf))
ans+=a;
cout<<ans<<endl;
}
}
