poj 1459 最大流(dinic实现) 

  1 #include <iostream>
  2 #include <cstring>
  3 
  4 #define N 1005
  5 #define M 50005
  6 #define I 0x7fffffff
  7 using namespace std;
  8 struct edge{
  9     int next, val, e;
 10 }net[M];
 11 int p[N], que[N], lev[N];
 12 int ind;
 13 inline void build (int s, int e, int w) {
 14     net[ind].e = e;
 15     net[ind].val = w;
 16     net[ind].next = p[s];
 17     p[s] = ind++;
 18     net[ind].e = s;
 19     net[ind].val = 0;
 20     net[ind].next = p[e];
 21     p[e] = ind++;
 22 }
 23 int dinic (int start, int end) {
 24     int t, path_n, k, i, h, r;
 25     int cur[M], path[M];
 26     int maxflow = 0;
 27     while (true) {
 28         memset (lev, 0, sizeof (lev) );
 29         h = r = 0;
 30         lev[start] = 1;
 31         que[0] = start;
 32         while (h <= r) {
 33             int t = que[h++];
 34             for (i = p[t];i != -1;i = net[i].next) {
 35                 if (net[i].val && lev[net[i].e] == 0) {
 36                     lev[net[i].e] = lev[t] + 1;
 37                     que[++r] = net[i].e;
 38                 }
 39             }
 40         }
 41         if (lev[end] == 0)
 42             break;
 43         memcpy (cur, p, sizeof (p) );
 44         path_n = -1;
 45         while (true) {
 46             if (path_n < 0) {
 47                 i = cur[start];
 48                 for (;i != -1;i = net[i].next) {
 49                     if (net[i].val && cur[net[i].e] != -1
 50                         && lev[net[i].e] == 2)
 51                         break;
 52                 }
 53                 if (i >= 0) {
 54                     path[++path_n] = i;
 55                     cur[start] = net[i].next;
 56                 }
 57                 else
 58                     break;
 59             }
 60             t = net[path[path_n]].e;
 61             if (t == end) {
 62                 int mink = -1;
 63                 int mmin = I;
 64                 for (k = 0;k <= path_n;k++) {
 65                     if (net[path[k]].val < mmin) {
 66                         mink = k;
 67                         mmin = net[path[k]].val;
 68                     }
 69                 }
 70                 maxflow += mmin;
 71                 for (k = 0;k <= path_n;k++) {
 72                     net[path[k]].val -= mmin;
 73                     net[path[k] ^ 1].val += mmin;
 74                 }
 75                 for (k = 0;k <= path_n;k++) {
 76                     if (net[path[k]].val == 0) {
 77                         path_n = mink - 1;
 78                         break;
 79                     }
 80                 }
 81             }
 82             else{
 83                 i = cur[t];
 84                 for (;i != -1;i = net[i].next) {
 85                     if (net[i].val && cur[net[i].e] != -1
 86                         && lev[net[i].e] == lev[t] + 1)
 87                         break;
 88                 }
 89                 if (i != -1) {
 90                     path[++path_n] = i;
 91                     cur[t] = net[i].next;
 92                 }
 93                 else{
 94                     cur[t] = -1;
 95                     path_n--;
 96                 }
 97             }
 98         }
 99     }
100     return maxflow;
101 }
102 int main (void)
103 {
104     int n,m,np,nc;
105     int flag[105];
106     int insum[105];
107     char tmp[30];
108     while (~scanf("%d%d%d%d",&n,&np,&nc,&m)) {
109         memset (p, -1, sizeof (p) );
110         ind = 0;
111         memset(flag,0,sizeof(flag));
112         memset(insum,0,sizeof(insum));
113         int u,v,z;
114         for(int i=0; i<m; i++)
115         {
116             scanf("%s",tmp);
117             sscanf(tmp,"(%d,%d)%d",&u,&v,&z);
118             u++,v++;
119             build(u+n,v,z);
120             insum[v] += z;
121         }
122 
123         for(int i=0; i<np; i++)
124         {
125             scanf("%s",tmp);
126             sscanf(tmp,"(%d)%d",&u,&z);
127             u++;
128             flag[u] = 1;
129             build(0,u,z);
130         }
131 
132         for(int i=0; i<nc; i++)
133         {
134             scanf("%s",tmp);
135             sscanf(tmp,"(%d)%d",&u,&z);
136             u++;
137             flag[u] = 2;
138             build(u+n,2*n+1,z);
139         }
140 
141         for(int i=1; i<=n; i++)
142             build(i,i+n,I);
143         printf("%d\n",dinic(0,2*n+1));
144     }
145     return 0;
146 }

 

posted @ 2014-04-18 21:52  辛力啤  阅读(186)  评论(0)    收藏  举报