POJ3204+DInic+maxflow

Dinic+maxflow
题意：找这样一种边的个数，就是增加该边的容量，可以使得最大流变大
思路：求maxflow，再枚举流量为0的边，增加容量，看是否能找到增广路径。

 

  1 /*
  2 Dinic+maxflow
  3 题意：找这样一种边的个数，就是增加该边的容量，可以使得最大流变大
  4 思路：求maxflow，再枚举流量为0的边，增加容量，看是否能找到增广路径。
  5 */
  6 #include<stdio.h>
  7 #include<string.h>
  8 #include<stdlib.h>
  9 #include<algorithm>
 10 #include<iostream>
 11 #include<queue>
 12 #include<map>
 13 #include<stack>
 14 #include<set>
 15 #include<math.h>
 16 using namespace std;
 17 typedef long long int64;
 18 //typedef __int64 int64;
 19 typedef pair<int64,int64> PII;
 20 #define MP(a,b) make_pair((a),(b)) 
 21 const int maxn = 505;
 22 const int maxm = 5005;
 23 const int inf = 0x3f3f3f3f;
 24 const double pi=acos(-1.0);
 25 const double eps = 1e-8;
 26 
 27 struct Edge{
 28     int u,v,next,val;
 29     bool flag;
 30 }edge[ maxm<<2 ];
 31 int cnt,head[ maxn ];
 32 int vis[ maxn ];
 33 int lev[ maxn ];
 34 int q[ maxn<<4 ];
 35 
 36 void init(){
 37     cnt = 0;
 38     memset( head,-1,sizeof( head ) );
 39 }
 40 void addedge( int a,int b,int c ){
 41     edge[ cnt ].u = a;
 42     edge[ cnt ].v = b;
 43     edge[ cnt ].val = c;
 44     edge[ cnt ].next = head[ a ];
 45     if( cnt%2==0 ) edge[ cnt ].flag = true;
 46     else edge[ cnt ].flag = false;
 47     head[ a ] = cnt ++;
 48 }
 49 
 50 bool bfs( int n,int start,int end ){
 51     int head2 = 0,tail2 = 0;
 52     q[ tail2++ ] = start;
 53     memset( lev,-1,sizeof( lev ) );
 54     lev[ start ] = 0;
 55     while( head2<tail2 ){
 56         int u = q[ head2++ ];
 57         for( int i=head[u];i!=-1;i=edge[i].next ){
 58             int v = edge[i].v;
 59             if( edge[i].val>0&&lev[v]==-1 ){
 60                 lev[v] = lev[u]+1;
 61                 q[ tail2++ ] = v;
 62             }
 63         }
 64     }
 65     if( lev[ end ]==-1 ) return false;
 66     else return true;
 67 }
 68 
 69 int Dinic( int n,int start,int end ){
 70     int maxflow = 0;
 71     while( true ){
 72         if( bfs(n,start,end )==false ) break;
 73         int id = start;
 74         int tail = 0;
 75         while( true ){
 76             if( id==end ){
 77                 int flow = inf;
 78                 int flag = -1;
 79                 for( int i=0;i<tail;i++ ){
 80                     if( edge[ q[i] ].val<flow ){
 81                         flow = edge[ q[i] ].val ;
 82                         flag = i;
 83                     }
 84                 }
 85                 for( int i=0;i<tail;i++ ){
 86                     edge[ q[i] ].val -= flow;
 87                     edge[ q[i]^1 ].val += flow;
 88                 }
 89                 if( flag!=-1 ){
 90                     maxflow += flow;
 91                     tail = flag;
 92                     id = edge[ q[flag] ].u;
 93                 }
 94                 else
 95                     return inf;
 96             }
 97             id = head[ id ];
 98             while( id!=-1 ){
 99                 if( edge[id].val>0&&lev[edge[id].u]+1==lev[edge[id].v] ){
100                     break;
101                 }
102                 id = edge[ id ].next;
103             }
104             if( id!=-1 ){
105                 q[ tail++ ] = id;
106                 id = edge[ id ].v;
107             }
108             else{
109                 if( tail==0 ) break;
110                 lev[ edge[q[tail-1]].v ] = -1;
111                 id = edge[ q[--tail] ].u;
112             }
113         }
114     }
115     return maxflow;
116 }                 
117 
118 int main(){
119     int n,m;
120     while( scanf("%d%d",&n,&m)==2 ){
121         init();
122         int a,b,c;
123         int start = 0;
124         int end = n-1;
125         for( int i=0;i<m;i++ ){
126             scanf("%d%d%d",&a,&b,&c);
127             addedge( a,b,c );
128             addedge( b,a,0 );
129         }
130         Dinic( n,start,end );
131         int ans = 0;
132         for( int i=0;i<cnt;i++ ){
133             if( edge[i].val==0&&edge[i].flag==true ){
134                 edge[i].val ++ ;
135                 if( bfs( n,start,end )==true ){
136                     ans ++ ;
137                 }
138                 edge[i].val -- ;
139             }
140         }
141         printf("%d\n",ans);
142     }
143     return 0;
144 }
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posted @ 2013-09-07 15:17  xxx0624  阅读(349)  评论(0)  编辑  收藏  举报