POJ1273 drainage ditches "max Stream" "SAP"

Topic Links:

http://poj.org/problem?id=1273

Main topic:

Farmer John's field has M ponds and n ditches for drainage, pond number 1~m,1 Pond is the source of all ditches,

Pond M is a ditch . meeting Point. give you the pond connected by n ditches and the amount of water that can flow, and ask the whole ditch from the source to the meeting point .

How much water can flow at most.

Ideas:

It is obvious that the maximum flow problem of network flow is obtained. Use a chain forward star (adjacency table) to store the network so you don't have to consider the problem of re-edge. It

In fact, the heavy edges are parallel edges. Using SAP Algorithm +GAP optimization to find the maximum flow. Sap+gap template refer to my other

A blog post: http://blog.csdn.net/lianai911/article/details/44962653

AC Code:

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <    queue>using namespace Std;const int maxn = 220;const int MAXM = maxn*maxn;const int INF = 0xffffff0;struct edgenode{    int to;    int W; int next;}    Edges[maxm];int head[maxn],id;void addedges (int u,int v,int w) {edges[id].to = v;    EDGES[ID].W = W;    Edges[id].next = Head[u];    Head[u] = id++;    edges[id].to = u;    EDGES[ID].W = 0;    Edges[id].next = Head[v]; HEAD[V] = id++;}    int numh[maxn],h[maxn],curedges[maxn],pre[maxn];void BFS (int end,int N) {memset (numh,0,sizeof (NUMH));    for (int i = 1; I <= N; ++i) numh[h[i]=n]++;    H[end] = 0;    numh[n]--;    numh[0]++;    Queue<int> Q;    Q.push (end); while (!        Q.empty ()) {int v = q.front ();        Q.pop ();        int i = head[v];            while (i! =-1) {int u = edges[i].to;                if (H[u] < N) {i = Edges[i].next; Continue           } H[u] = H[v] + 1;            numh[n]--;            numh[h[u]]++;            Q.push (U);        i = Edges[i].next;    }}}int SAP (int start,int end,int N) {int curflow,flowans = 0,temp,neck;    memset (h,0,sizeof (h));    memset (pre,-1,sizeof (pre));    for (int i = 1; I <= N; ++i) curedges[i] = Head[i];    BFS (End,n);    int u = start;            while (H[start] < N) {if (U = = end) {Curflow = INF;                for (int i = start; I! = end; i = edges[curedges[i]].to) {if (Curflow > EDGES[CUREDGES[I]].W)                    {neck = i;                Curflow = EDGES[CUREDGES[I]].W; }} for (int i = start; I! = end; i = edges[curedges[i]].to) {temp = cured                Ges[i];                EDGES[TEMP].W-= Curflow;            EDGES[TEMP^1].W + = Curflow;            } Flowans + = Curflow;        u = neck;  }      int i;                for (i = curedges[u]; i =-1; i = edges[i].next) if (EDGES[I].W && h[u]==h[edges[i].to]+1)        Break            if (i! =-1) {curedges[u] = i;            Pre[edges[i].to] = u;        u = edges[i].to;            } else {if (0 = =--numh[h[u]]) break;            Curedges[u] = Head[u]; for (temp = N,i = Head[u]; I! =-1; i = edges[i].next) if (EDGES[I].W) temp = min (temp,h[e            Dges[i].to]);            H[u] = temp + 1;            ++numh[h[u]];        if (u! = start) U = pre[u]; }} return Flowans;}    int main () {int n,m,u,v,w;        while (~SCANF ("%d%d", &n,&m)) {memset (head,-1,sizeof (Head));        id = 0;            for (int i = 0; i < N; ++i) {scanf ("%d%d%d", &u,&v,&w);        Addedges (U,V,W);    } printf ("%d\n", SAP (1,m,m)); } return 0;}

POJ1273 drainage ditches "max Stream" "SAP"