最大流sap模板 （转自庄神）

最大流sap模板 

转自庄神：Orz zsO  

以后就用它了 好高兴。。。233333

http://www.zlinkin.com/?p=34

typedef  struct {    int v, next, val;} edge;const int MAXN = 20010;const int MAXM = 500010; edge e[MAXM];int p[MAXN], eid; inline void init() {    memset(p, -1, sizeof(p));    eid = 0;} //有向inline void insert1(int from, int to, int val) {    e[eid].v = to;    e[eid].val = val;    e[eid].next = p[from];    p[from] = eid++;     swap(from, to);     e[eid].v = to;    e[eid].val = 0;    e[eid].next = p[from];    p[from] = eid++;} //无向inline void insert2(int from, int to, int val) {    e[eid].v = to;    e[eid].val = val;    e[eid].next = p[from];    p[from] = eid++;     swap(from, to);     e[eid].v = to;    e[eid].val = val;    e[eid].next = p[from];    p[from] = eid++;} int n, m; //n为点数 m为边数int h[MAXN];int gap[MAXN]; int source, sink;inline int dfs(int pos, int cost) {    if (pos == sink) {        return cost;    }     int j, minh = n - 1, lv = cost, d;     for (j = p[pos]; j != -1; j = e[j].next) {        int v = e[j].v, val = e[j].val;         if(val > 0) {            if (h[v] + 1 == h[pos]) {                if (lv < e[j].val) {                    d = lv;                } else {                    d = e[j].val;                }                 d = dfs(v, d);                e[j].val -= d;                e[j ^ 1].val += d;                lv -= d;                 if (h[source] >= n) {                    return cost - lv;                }                 if (lv == 0) {                    break;                }            }             if (h[v] < minh){                minh = h[v];            }        }    }     if (lv == cost) {        --gap[h[pos]];         if (gap[h[pos]] == 0) {            h[source] = n;        }         h[pos] = minh + 1;        ++gap[h[pos]];    }     return cost - lv; } int sap(int st, int ed) {     source = st;    sink = ed;    int ret = 0;    memset(gap, 0, sizeof(gap));    memset(h, 0, sizeof(h));     //gap[st] = n;    gap[0] = n;     while (h[st] < n) {        ret += dfs(st, INT_MAX);    }     return ret;}