【二分+最小树形图】UVA11865 比赛网络
Description
During 2009 and 2010 ICPC world finals, the contest was webcasted via world wide web. Seeing this, some contest organizers from Ajobdesh decided that, they will provide a live stream of their contests to every university in Ajobdesh. The organizers have decided that, they will provide best possible service to them. But there are two problems: 1. There is no existing network between universities. So, they need to build a new network. However, the maximum amount they can spend on building the network is C. 2. Each link in the network has a bandwidth. If, the stream’s bandwidth exceeds any of the link’s available bandwidth, the viewers, connected through that link can’t view the stream. Due to the protocols used for streaming, a viewer can receive stream from exactly one other user (or the server, where the contest is organized). That is, if you have two 128kbps links, you won’t get 256kbps bandwidth, although, if you have a stream of 128kbps, you can stream to any number of users at that bandwidth. Given C, you have to maximize the minimum bandwidth to any user.
Solution
二分最小带宽，求最小树形图看是否超过C。
Code
写这题各种犯逗，简直感动不能多说。
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 using namespace std; 5 const int maxn=65,maxm=1e4+5; 6 7 struct edge{ 8 int v,u,b,c; 9 bool operator<(const edge&a) 10 const{return b<a.b;} 11 }E[maxm],e[maxm]; 12 int in[maxn],pre[maxn],vis[maxn],id[maxn]; 13 int N,M,C; 14 15 int work(int n,int m,int lim){ 16 memcpy(e,E,sizeof(e)); 17 long long ret=0; 18 int root=1; 19 if(!lim) lim=1; 20 21 while(1){ 22 memset(in,127,sizeof(in)); 23 int inf=in[0]; 24 for(int i=lim;i<=m;i++){ 25 int u=e[i].u,v=e[i].v; 26 if(u!=v&&e[i].c<in[v]){ 27 in[v]=e[i].c; 28 pre[v]=u; 29 } 30 } 31 for(int i=1;i<=n;i++) 32 if(i!=root&&in[i]==inf) return 0; 33 34 in[root]=0; 35 int cnt=0; 36 memset(id,0,sizeof(id)); 37 memset(vis,0,sizeof(vis)); 38 for(int i=1;i<=n;i++){ 39 ret+=in[i]; 40 if(!vis[i]){ 41 int u=i; 42 while(u!=root&&!vis[u]){ 43 vis[u]=i; 44 u=pre[u]; 45 } 46 if(vis[u]==i){ 47 ++cnt; 48 int v=u; 49 do{ 50 id[v]=cnt; 51 v=pre[v]; 52 }while(v!=u); 53 } 54 } 55 } 56 57 if(!cnt) break; 58 for(int i=1;i<=n;i++) 59 if(!id[i]) id[i]=++cnt; 60 for(int i=lim;i<=m;i++){ 61 int v=e[i].v; 62 e[i].u=id[e[i].u]; 63 e[i].v=id[e[i].v]; 64 if(e[i].u!=e[i].v) 65 e[i].c-=in[v]; 66 } 67 n=cnt; 68 root=id[root]; 69 } 70 if(ret<=C) return 1; 71 return 0; 72 } 73 74 int main(){ 75 int T; 76 scanf("%d",&T); 77 78 while(T--){ 79 scanf("%d%d%d",&N,&M,&C); 80 for(int i=1;i<=M;i++){ 81 scanf("%d%d%d%d",&E[i].u,&E[i].v,&E[i].b,&E[i].c); 82 E[i].u++,E[i].v++; 83 } 84 sort(E+1,E+M+1); 85 86 int l=0,r=M; 87 while(l<r){ 88 int mid=(l+r+1)>>1; 89 if(work(N,M,mid)) l=mid; 90 else r=mid-1; 91 } 92 93 if(!l) printf("streaming not possible.\n"); 94 else printf("%d kbps\n",E[l].b); 95 } 96 return 0; 97 }
