題目鏈接：點(diǎn)擊查看

題目大意：給出一個 n 個點(diǎn)和 m 條邊的有向圖，每條邊都有一個流量限制 [ lower , upper ]，給定源點(diǎn) s 和匯點(diǎn) t ，求出源點(diǎn)到匯點(diǎn)的最小流

題目分析：參考我的上一篇博客：https://blog.csdn.net/qq_45458915/article/details/108339354

坑點(diǎn)就是卡常+卡當(dāng)前弧優(yōu)化，第八個點(diǎn)會TLE，如果有當(dāng)前弧優(yōu)化可以刪除試一下，如果鏈?zhǔn)角跋蛐鞘怯媒Y(jié)構(gòu)體封裝的可以解封裝試一下

代碼：

#include<iostream> #include<cstdio> #include<string> #include<ctime> #include<cmath> #include<cstring> #include<algorithm> #include<stack> #include<climits> #include<queue> #include<map> #include<set> #include<sstream> #include<cassert> #include<bitset> #include<unordered_map> using namespace std;typedef long long LL;typedef unsigned long long ull;const int inf=0x3f3f3f3f;const int N=50100;const int M=125100;int du[N];int head[N],cnt,edge[(N+M)<<1],ver[(N+M)<<1],nt[(N+M)<<1];void addedge(int u,int v,int w) {ver[cnt]=v;edge[cnt]=w;nt[cnt]=head[u];head[u]=cnt++;ver[cnt]=u;edge[cnt]=0;//反向邊邊權(quán)設(shè)置為0nt[cnt]=head[v];head[v]=cnt++; }int d[N];//深度bool bfs(int s,int t)//尋找增廣路 {memset(d,0,sizeof(d));queue<int>q;q.push(s);d[s]=1;while(!q.empty()){int u=q.front();q.pop();for(int i=head[u];i!=-1;i=nt[i]){if(d[ver[i]])continue;if(!edge[i])continue;d[ver[i]]=d[u]+1;q.push(ver[i]);if(ver[i]==t)return true;}}return false; }int dinic(int x,int t,int flow)//更新答案 {if(x==t)return flow;int rest=flow,i;for(i=head[x];i!=-1&&rest;i=nt[i]){if(edge[i]&&d[ver[i]]==d[x]+1){int k=dinic(ver[i],t,min(rest,edge[i]));if(!k)d[ver[i]]=0;edge[i]-=k;edge[i^1]+=k;rest-=k;}}return flow-rest; }void init() {memset(head,-1,sizeof(head));cnt=0; }int solve(int st,int ed) {int ans=0,flow;while(bfs(st,ed))while(flow=dinic(st,ed,inf))ans+=flow;return ans; }int main() { #ifndef ONLINE_JUDGE // freopen("data.in.txt","r",stdin); // freopen("data.out.txt","w",stdout); #endif // ios::sync_with_stdio(false);init();int n,m,s,t;scanf("%d%d%d%d",&n,&m,&s,&t);while(m--){int u,v,lower,upper;scanf("%d%d%d%d",&u,&v,&lower,&upper);addedge(u,v,upper-lower);du[u]-=lower,du[v]+=lower;}int st=N-1,ed=st-1,sum=0;for(int i=1;i<=n;i++){if(du[i]>0){addedge(st,i,du[i]);sum+=du[i];}elseaddedge(i,ed,-du[i]);}addedge(t,s,inf);if(solve(st,ed)!=sum){puts("please go home to sleep");return 0;}int ans=edge[cnt-1];edge[cnt-1]=edge[cnt-2]=0;printf("%d\n",ans-solve(t,s));return 0; }

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