Drainage Ditches
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 68414 | Accepted: 26487 |
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.
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.
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
网络流最大流经典入门题，学习了算法竞赛入门经典的程序，我用两种方法解，即Dinic算法和ISAP算法。
Dinic算法（邻接表、无cur优化）
15705294
| ksq2013 | 1273 | Accepted | 732K | 0MS | G++ | 1634B | 2016-07-11 20:44:19 |
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int INF=0x3f3f3f3f;
int n,m,s,t,nxt[800],first[800],ecnt;
struct Edge{int u,v,cap,flow;}e[800];
bool vis[800];
int d[800],cur[800];
int bfs()
{
memset(vis,false,sizeof(vis));
queue<int>q;
q.push(s);
d[s]=0;
vis[s]=true;
while(!q.empty()){
int now=q.front();q.pop();
for(int i=first[now];i;i=nxt[i]){
if(!vis[e[i].v]&&e[i].cap>e[i].flow){
vis[e[i].v]=true;
d[e[i].v]=d[now]+1;
q.push(e[i].v);
}
}
}
return vis[t];
}
int dfs(int x,int a)
{
if(x==t||a==0)return a;
int flow=0,f;
for(int i=first[x];i;i=nxt[i])
if(d[e[i].v]==d[x]+1&&(f=dfs(e[i].v,min(a,e[i].cap-e[i].flow)))>0){
e[i].flow+=f;
e[i^1].flow-=f;
flow+=f;
a-=f;
if(a==0)break;
}
return flow;
}
int Dinic()
{
int flow=0;
while(bfs()){
memset(cur,0,sizeof(cur));
flow+=dfs(s,INF);
}
return flow;
}
void Link()
{
memset(nxt,0,sizeof(nxt));
memset(first,0,sizeof(first));
for(int a,b,c;m;m--){
scanf("%d%d%d",&a,&b,&c);
e[++ecnt].u=a,e[ecnt].v=b,e[ecnt].cap=c,e[ecnt].flow=0;
nxt[ecnt]=first[a],first[a]=ecnt;
e[++ecnt].u=b,e[ecnt].v=a,e[ecnt].cap=0,e[ecnt].flow=0;
nxt[ecnt]=first[b],first[b]=ecnt;
}
}
int main()
{
while(~scanf("%d%d",&m,&n)){
s=1,t=n,ecnt=1;
memset(d,0,sizeof(d));
Link();
printf("%d\n",Dinic());
}
return 0;
}
ISAP算法（邻接表，有gap等优化）
15708393
| ksq2013 | 1273 | Accepted | 736K | 16MS | G++ | 2300B | 2016-07-12 10:59:51 |
#include<cstdio>
#include<cstring>
#include<iostream>
#define INF 0x3f3f3f3f
using namespace std;
int n,m,s,t,ecnt,first[800],nxt[800];
struct Edge{int u,v,cap,flow;}e[800];
bool vis[800];
int q[800],d[800],p[800],num[800],cur[800];
void Link()
{
memset(first,0,sizeof(first));
memset(nxt,0,sizeof(nxt));
for(int a,b,c;m;m--){
scanf("%d%d%d",&a,&b,&c);
e[++ecnt].u=a,e[ecnt].v=b,e[ecnt].cap=c,e[ecnt].flow=0;
nxt[ecnt]=first[e[ecnt].u];first[e[ecnt].u]=ecnt;
e[++ecnt].u=b,e[ecnt].v=a,e[ecnt].cap=0,e[ecnt].flow=0;
nxt[ecnt]=first[e[ecnt].u];first[e[ecnt].u]=ecnt;
}
}
void bfs()
{
memset(vis,false,sizeof(vis));
int head=0,tail=1;
q[0]=t;
d[t]=0;
vis[t]=true;
while(head^tail){
int now=q[head];head++;
for(int i=first[now];i;i=nxt[i])
if(!vis[e[i].u]&&e[i].cap>e[i].flow){
vis[e[i].u]=true;
d[e[i].u]=d[now]+1;
q[tail++]=e[i].u;
}
}
}
int Agument()
{
int x=t,a=INF;
while(x^s){
a=min(a,e[p[x]].cap-e[p[x]].flow);
x=e[p[x]].u;
}
x=t;
while(x^s){
e[p[x]].flow+=a;
e[p[x]^1].flow-=a;
x=e[p[x]].u;
}
return a;
}
int ISAP()
{
int flow=0;
bfs();
memset(num,0,sizeof(num));
for(int i=1;i<=n;i++)num[d[i]]++;
int x=s;
for(int i=1;i<=n;i++)cur[i]=first[i];//memset(cur,0,sizeof(cur));
while(d[s]<n){
if(!(x^t)){
flow+=Agument();
x=s;
}
bool advanced=false;
for(int i=cur[x];i;i=nxt[i])
if(e[i].cap>e[i].flow&&d[x]==d[e[i].v]+1){
advanced=true;
p[e[i].v]=i;
cur[x]=i;
x=e[i].v;
break;
}
if(!advanced){
int mn=n-1;
for(int i=first[x];i;i=nxt[i])
if(e[i].cap>e[i].flow)mn=min(mn,d[e[i].v]);
if(--num[d[x]]==0)break;
num[d[x]=mn+1]++;
cur[x]=first[x];
if(x^s)x=e[p[x]].u;
}
}
return flow;
}
int main()
{
while(~scanf("%d%d",&m,&n)){
s=1,t=n,ecnt=1;
Link();
memset(d,0,sizeof(d));
memset(p,0,sizeof(p));
printf("%d\n",ISAP());
}
return 0;
}