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hdu 3879 方案选择

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 1 #include <cstdio>
 2 #include <cstring>
 3 #include <vector>
 4 #define oo 0x3f3f3f3f
 5 #define N 55010
 6 using namespace std;
 7 
 8 struct Edge {
 9     int u, v, f;
10     Edge( int u, int v, int f ):u(u),v(v),f(f){}
11 };
12 
13 int n, m, src, dst;
14 vector<Edge> edge;
15 vector<int> g[N];
16 int dep[N], cur[N], qu[N], bg, ed, idc;
17 int sump;
18 
19 void init() {
20     for( int i=1; i<=n+m+2; i++ )
21         g[i].clear();
22     edge.clear();
23     sump = 0;
24     idc = 0;
25 }
26 void adde( int u, int v, int f ) {
27     g[u].push_back( edge.size() );
28     edge.push_back( Edge(u,v,f) );
29     g[v].push_back( edge.size() );
30     edge.push_back( Edge(v,u,0) );
31 }
32 bool bfs() {
33     memset( dep, 0, sizeof(dep) );
34     qu[bg=ed=1] = src;
35     dep[src] = 1;
36     while( bg<=ed ) {
37         int u=qu[bg++];
38         for( int t=0; t<g[u].size(); t++ ) {
39             Edge &e = edge[g[u][t]];
40             if( e.f && !dep[e.v] ) {
41                 dep[e.v] = dep[e.u] + 1;
42                 qu[++ed] = e.v;
43             }
44         }
45     }
46     return dep[dst];
47 }
48 int dfs( int u, int a ) {
49     if( u==dst || a==0 ) return a;
50     int remain=a, past=0, na;
51     for( int &t=cur[u]; t<g[u].size(); t++ ) {
52         Edge &e = edge[g[u][t]];
53         Edge &ve = edge[g[u][t]^1];
54         if( e.f && dep[e.v]==dep[e.u]+1 && (na=dfs(e.v,min(remain,e.f))) ) {
55             remain -= na;
56             past += na;
57             e.f -= na;
58             ve.f += na;
59             if( !remain ) break;
60         }
61     }
62     return past;
63 }
64 int maxflow() {
65     int flow = 0;
66     while( bfs() ) {
67         memset( cur, 0, sizeof(cur) );
68         flow += dfs(src,oo);
69     }
70     return flow;
71 }
72 int main() {
73     while( scanf( "%d%d", &n, &m )==2 ) {
74         init();
75         src = n+m+1;
76         dst = src+1;
77         for( int i=1,c; i<=n; i++ ) {
78             scanf( "%d", &c );
79             ++idc;
80             adde( idc, dst, c );
81         }
82         for( int i=1,u,v,p; i<=m; i++ ) {
83             scanf( "%d%d%d", &u, &v, &p );
84             ++idc;
85             sump += p;
86             adde( src, idc, p );
87             adde( idc, u, oo );
88             adde( idc, v, oo );
89         }
90         printf( "%d\n", sump-maxflow() );
91     }
92 }
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hdu 3879 方案选择

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原文地址:http://www.cnblogs.com/idy002/p/4531109.html

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