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sgu 194 上下界网络流可行流判定+输出可行流

时间:2015-06-03 22:59:10      阅读:118      评论:0      收藏:0      [点我收藏+]

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  1 #include <cstdio>
  2 #include <cstring>
  3 #define min(a,b) ((a)<(b)?(a):(b))
  4 #define oo 0x3f3f3f3f
  5 #define N 210
  6 #define M 100010
  7 
  8 struct Dinic {
  9     int n, src, dst;
 10     int head[N], dest[M], flow[M], eid[M], next[M], etot;
 11     int cur[N], dep[N], qu[N], bg, ed;
 12 
 13     void init( int n, int src, int dst ) {
 14         this->n = n;
 15         this->src = src;
 16         this->dst = dst;
 17         etot = 0;
 18         memset( head, -1, sizeof(head) );
 19     }
 20     void adde( int id, int u, int v, int f ) {
 21         next[etot]=head[u], flow[etot]=f, dest[etot]=v, eid[etot]=id; head[u]=etot++;
 22         next[etot]=head[v], flow[etot]=0, dest[etot]=u, eid[etot]=-1; head[v]=etot++;
 23     }
 24     bool bfs() {
 25         memset( dep, 0, sizeof(dep) );
 26         qu[bg=ed=1] = src;
 27         dep[src] = 1;
 28         while( bg<=ed ) {
 29             int u=qu[bg++];
 30             for( int t=head[u]; ~t; t=next[t] ) {
 31                 int v=dest[t], f=flow[t];
 32                 if( f && !dep[v] ) {
 33                     dep[v] = dep[u]+1;
 34                     qu[++ed] = v;
 35                 }
 36             }
 37         }
 38         return dep[dst];
 39     }
 40     int dfs( int u, int a ) {
 41         if( u==dst || a==0 ) return a;
 42         int remain=a, past=0, na;
 43         for( int &t=cur[u]; ~t; t=next[t] ) {
 44             int v=dest[t], &f=flow[t], &vf=flow[t^1];
 45             if( f && dep[v]==dep[u]+1 && (na=dfs(v,min(remain,f))) ) {
 46                 f -= na;
 47                 vf += na;
 48                 remain -= na;
 49                 past += na;
 50                 if( !remain ) break;
 51             }
 52         }
 53         return past;
 54     }
 55     int maxflow() {
 56         int f = 0;
 57         while( bfs() ) {
 58             for( int u=1; u<=n; u++ ) cur[u]=head[u];
 59             f += dfs(src,oo);
 60         }
 61         return f;
 62     }
 63     void print( int *ans ) {
 64         for( int e=0; e<etot; e++ )
 65             if( eid[e]!=-1 ) ans[eid[e]]+=flow[e^1];
 66     }
 67 }D;
 68 struct Bottop {
 69     int n;
 70     int head[N], dest[M], bot[M], top[M], next[M], eid[M], etot;
 71     int si[N], so[N], sum;
 72 
 73     void init( int n ) {
 74         this->n = n;
 75         memset( head, -1, sizeof(head) );
 76         memset( si, 0, sizeof(si) );
 77         memset( so, 0, sizeof(so) );
 78         etot = 0;
 79     }
 80     void adde( int id, int u, int v, int b, int t ) {
 81         eid[etot]=id, next[etot]=head[u], bot[etot]=b, top[etot]=t, dest[etot]=v; 
 82         head[u]=etot++;
 83     }
 84     void build( int *ans ) {
 85         int src=n+1, dst=n+2;
 86         D.init( dst, src, dst );
 87         for( int u=1; u<=n; u++ ) 
 88             for( int t=head[u]; ~t; t=next[t] ) {
 89                 int v=dest[t];
 90                 ans[eid[t]] += bot[t];
 91                 si[v] += bot[t];
 92                 so[u] += bot[t];
 93                 D.adde( eid[t], u, v, top[t]-bot[t] );
 94             }
 95         for( int u=1; u<=n; u++ ) {
 96             if( so[u]>si[u] ) 
 97                 D.adde( -1, u, dst, so[u]-si[u] );
 98             else if( si[u]>so[u] ) {
 99                 D.adde( -1, src, u, si[u]-so[u] );
100                 sum += si[u]-so[u];
101             }
102         }
103     }
104     bool ok() { return D.maxflow()==sum; }
105 }B;
106 
107 int n, m;
108 int ans[M];
109 
110 int main() {
111     scanf( "%d%d", &n, &m );
112     B.init(n);
113     for( int i=1,u,v,b,t; i<=m; i++ ) {
114         scanf( "%d%d%d%d", &u, &v, &b, &t );
115         B.adde( i, u, v, b, t );
116     }
117     B.build(ans);
118     bool ok = B.ok();
119     printf( "%s\n", ok ? "YES" : "NO" );
120     if( ok ) {
121         D.print( ans );
122         for( int i=1; i<=m; i++ )
123             printf( "%d\n", ans[i] );
124     }
125 }
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sgu 194 上下界网络流可行流判定+输出可行流

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原文地址:http://www.cnblogs.com/idy002/p/4550323.html

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