POJ 3228 Gold Transportation two points + max flow

http://poj.org/problem?id=3228

Test instructions: There are n cities, each city has a certain amount of gold, we need to put all the gold in the city's warehouses, there are some roads connected to these cities, each road has its own cost, the demand is that all the gold is transported to the warehouse, the road to go, where the maximum cost of the minimum is how much.

Thinking, two-point answer, the cost is less than the answer of those roads to build a diagram, and then establish a source point and all cities connected, traffic is the number of gold per city, and then set up a meeting point, connecting all cities, traffic is the maximum amount of gold in each city. Run the maximum flow again, if the flow equals all the gold, then you can, otherwise you can't.

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace Std;


const int INF=1E9;
struct side{
int from,to,val;
}SIDE1[20010];
struct side2{
int to,val,next;
}SIDE2[40010];
int n,m;
int a[210],b[210];
int cnt,dis[210],gap[210];
int node[210],top;
int sum;


void addside (int a,int b,int c) {
Side2[top]= (Side2) {b,c,node[a]};
node[a]=top++;
Side2[top]= (Side2) {a,0,node[b]};
node[b]=top++;
}


int getflow (int u,int f) {
if (u==n+1)
return F;
int ans=0;
for (int i=node[u];i!=-1;i=side2[i].next) {
int to=side2[i].to,c=side2[i].val;
if (dis[u]>dis[to]&&c) {
int X=getflow (To,min (f-ans,c));
Side2[i].val-=x;
Side2[i^1].val+=x;
Ans+=x;
if (ans==f) return ans;
}
}
gap[dis[u]]--;
if (gap[dis[u]]==0) dis[0]=cnt+2;
dis[u]++;
gap[dis[u]]++;
return ans;
}


bool OK (int x) {
cnt=n+2;
memset (dis,0,sizeof (dis));
memset (Gap,0,sizeof (GAP));
memset (node,-1,sizeof (node));
gap[0]=cnt;
Top=0;
for (int i=1;i<=n;i++) {
Addside (0,i,a[i]);
Addside (I,n+1,b[i]);
}
for (int i=0;i<m;i++) if (side1[i].val<=x) {
Addside (Side1[i].from,side1[i].to,inf);
Addside (Side1[i].to,side1[i].from,inf);
}
int ans=0;
while (dis[0]<cnt) Ans+=getflow (0,inf);
if (ans==sum)
return 1;
else return 0;
}




int main ()
{
while (Cin>>n) {
sum=0;
if (n==0) break;
for (int i=1;i<=n;i++) {
scanf ("%d", &a[i]);
Sum+=a[i];
}
for (int i=1;i<=n;i++)
scanf ("%d", &b[i]);
cin>>m;
int x, y, Z;
int l=1e9,r=0,mid;
for (int i=0;i<m;i++) {
scanf ("%d%d%d", &x,&y,&z);
Side1[i]= (Side) {x, y, z};
L=min (L,Z);
R=max (R,Z);
}
r++;
int maxx=r;
while (l!=r) {
Mid= (L+R)/2;
if (OK (mid))
R=mid;
else l=mid+1;
}
if (L==maxx)
cout<< "No Solution" <<endl;
else cout<<l<<endl;
}
return 0;
}

POJ 3228 Gold Transportation two points + max flow