POJ 1459 Power Network (multi-source/sink max flow problem)

Title Link: http://poj.org/problem?id=1459

The topic gives you a big paragraph of explanation, is actually the nonsense. It is misleading to give an explanatory picture.

The main idea: for a grid, there are power stations, electricity, and transmission lines. The power station is limited, the power supply is limited, the transmission line is limited, so obviously a network flow problem. First give the line and the limit, then give the electricity side, and finally out of the power station.

Because it is a multi-source point (multiple power stations), a multi-meeting point (multiple power consumption), so the need for super-source processing.

As a super-source, it is assumed that there is a source, connected to all the source point (Power station), its line capacity is the limit of the power station, then the power station can be treated as a normal point. Assuming a super meeting point, then we can connect all the meeting point (electricity) to this super meeting point, its line capacity is the power-side limit, then, it becomes a simple single source point, single sink point maximum flow problem. It can be solved with dinic.

#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <queue > #include <algorithm> #define MAX 999999using namespace Std;int map_[200][200];int dis[200];int bfs (int s,int t)    {int now;    memset (dis,-1,sizeof (dis));    Dis[s] = 0;    Queue<int> que;    Que.push (s);        while (!que.empty ()) {now = Que.front ();        Que.pop (); for (int i = 0;i <= t;i++) if (dis[i] = = 1 && map_[now][i] > 0) {dis[                I] = Dis[now] + 1;            Que.push (i);    }} if (Dis[t]! =-1) return 1; return 0;}    int dinic (int s,int t,int x) {if (s = = t) return x;    int tmp = x; for (int i = 0;i <= t;i++) {if (dis[i] = = Dis[s] + 1 && map_[s][i] > 0) {int i            min = Dinic (i,t,min (map_[s][i],x));            Map_[s][i]-= imin;            Map_[i][s] + = Imin;        X-= Imin; }} return tmp -X;}    int main () {int n,np,nc,m;        while (~SCANF ("%d%d%d%d", &n,&np,&nc,&m)) {int i,k;        int u,v,c;        memset (map_,0,sizeof (MAP_));            for (i = 0;i < m;i++) {scanf ("(%d,%d)%d", &u,&v,&c);    Map_[u + 1][v + 1] + = C;            0 is the super source point, the other point is moved back} for (i = 0;i < np;i++) {scanf ("(%d)%d", &v,&c);        Map_[0][v + 1] + = C;            } for (i = 0;i < nc;i++) {scanf ("(%d)%d", &u,&c);        Map_[u + 1][n + 1] + = C;        } int ans = 0;        while (BFS (0,n + 1)) ans + = Dinic (0,n + 1,max);    printf ("%d\n", ans); } return 0;}

do not forget to beginner's mind, the party must always

POJ 1459 Power Network (multi-source/sink max flow problem)