sgu 176 上下界网络流最小可行流带输出方案

  1 #include <cstdio>
  2 #include <cassert>
  3 #include <cstring>
  4 #define min(a,b) ((a)<(b)?(a):(b))
  5 #define oo 0x3f3f3f3f
  6 #define N 110
  7 #define M N*N
  8 
  9 struct Dinic {
 10     int n, src, dst;
 11     int head[N], dest[M], flow[M], next[M], info[M], etot;
 12     int cur[N], dep[N], qu[N], bg, ed;
 13     void init( int n ) {
 14         this->n = n;
 15         memset( head, -1, sizeof(head) );
 16         etot = 0;
 17     }
 18     void adde( int i, int u, int v, int f ) {
 19         info[etot]=i, flow[etot]=f, dest[etot]=v, next[etot]=head[u]; head[u]=etot++;
 20         info[etot]=0, flow[etot]=0, dest[etot]=u, next[etot]=head[v]; head[v]=etot++;
 21     }
 22     bool bfs() {
 23         memset( dep, 0, sizeof(dep) );
 24         qu[bg=ed=1] = src;
 25         dep[src] = 1;
 26         while( bg<=ed ) {
 27             int u=qu[bg++];
 28             for( int t=head[u]; t!=-1; t=next[t] ) {
 29                 int v=dest[t], f=flow[t];
 30                 if( f && !dep[v] ) {
 31                     dep[v]=dep[u]+1;
 32                     qu[++ed] = v;
 33                 }
 34             }
 35         }
 36         return dep[dst];
 37     }
 38     int dfs( int u, int a ) {
 39         if( u==dst || a==0 ) return a;
 40         int remain=a, past=0, na;
 41         for( int &t=cur[u]; t!=-1; t=next[t] ) {
 42             int v=dest[t], &f=flow[t], &vf=flow[t^1];
 43             if( f && dep[v]==dep[u]+1 && (na=dfs(v,min(remain,f))) ) {
 44                 f -= na;
 45                 vf += na;
 46                 remain -= na;
 47                 past += na;
 48                 if( !remain ) break;
 49             }
 50         }
 51         return past;
 52     }
 53     int maxflow( int s, int t ) {
 54         int f = 0;
 55         src = s, dst = t;
 56         while( bfs() ) {
 57             memcpy( cur, head, sizeof(cur) );
 58             f += dfs(src,oo);
 59         }
 60         return f;
 61     }
 62 }dinic;
 63 struct Btop {
 64     int n;
 65     int head[N], dest[M], bval[M], tval[M], next[M], info[M], etot;
 66     int sumi[N], sumo[N];
 67     void init( int n ) {
 68         etot = 0;
 69         memset( head, -1, sizeof(head) );
 70         this->n = n;
 71     }
 72     void adde( int i, int u, int v, int b, int t ) {
 73         info[etot]=i, bval[etot]=b, tval[etot]=t;
 74         dest[etot]=v, next[etot]=head[u];
 75         sumi[v]+=b, sumo[u]+=b;
 76         head[u] = etot++;
 77     }
 78     int minflow( int src, int dst ) {
 79         int ss=n+1, tt=n+2, sum;
 80         dinic.init( n+2 );
 81         for( int u=1; u<=n; u++ )
 82             for( int t=head[u]; t!=-1; t=next[t] ) {
 83                 int v=dest[t];
 84                 dinic.adde( info[t], u, v, tval[t]-bval[t] );
 85             }
 86         sum = 0;
 87         for( int u=1; u<=n; u++ ) {
 88             if( sumi[u]>sumo[u] ) {
 89                 dinic.adde( 0, ss, u, sumi[u]-sumo[u] );
 90                 sum += sumi[u]-sumo[u];
 91             } else if( sumo[u]>sumi[u] ) {
 92                 dinic.adde( 0, u, tt, sumo[u]-sumi[u] );
 93             }
 94         }
 95         int f = 0;
 96         f += dinic.maxflow(ss,tt);
 97         dinic.adde( 0, dst, src, oo );
 98         f += dinic.maxflow(ss,tt);
 99         if( f!=sum ) return -1;
100         int eid = dinic.etot-2;
101         return dinic.flow[eid^1];
102     }
103 }btop;
104 
105 int n, m;
106 int ans[M], tot;
107 
108 int main() {
109     scanf( "%d%d", &n, &m );
110     btop.init( n );
111     for( int i=1,u,v,z,c; i<=m; i++ ) {
112         scanf( "%d%d%d%d", &u, &v, &z, &c );
113         if( c==1 ) btop.adde( i, u, v, z, z );
114         else btop.adde( i, u, v, 0, z );
115         ans[i] = c ? z : 0;
116     }
117     int minf = btop.minflow(1,n);
118     if( minf==-1 ) {
119         printf( "Impossible\n" );
120         return 0;
121     }
122     for( int e=0; e<dinic.etot; e++ ) {
123         int i=dinic.info[e];
124         if( i ) ans[i] += dinic.flow[e^1];
125     }
126     printf( "%d\n", minf );
127     for( int i=1; i<=m; i++ )
128         printf( "%d ", ans[i] );
129     printf( "\n" );
130 }