bzoj-3931 Network throughput

Test instructions

A graph of n nodes is given, and each node has a throughput;

The rule is that the packet delivery must follow the shortest path, and the network throughput of 1 to n is obtained;

Exercises
As Huang said, the problem is test instructions;

But there are a few things to be aware of;

The shortest circuit is dij or SPFA, but for the purpose of building a map, you need to record the path;

If the chain forward star can be all used to update the side to save the map;

But using vectors or adjacency matrices (not knowing if they can live) can not be achieved;

I use a vector array to hang the chain above;

Topology queue sweep over plus side, the complexity is quite must not;

Then the demolition of the map is nothing to say;

The shortest circuit and flow are to open long long, the throughput of the first and last point is not counted;

Code:

#include <queue> #include <vector> #include <stdio.h> #include <string.h> #include < Algorithm> #define PR pair<int,int> #define N 600#define M 210000using namespace Std;typedef long Long ll;queue&lt ; INT&GT;Q;PRIORITY_QUEUE&LT;PR,VECTOR&LT;PR &GT;,GREATER&LT;PR > >poq;vector<int>to[n],pre[n];vector <ll>val[n];int head[n<<1],next[m],t[m],n,tot=1;ll dis[n<<1],flow[m];bool vis[N];void Add (int x, int y,ll fl) {T[++TOT]=Y,NEXT[TOT]=HEAD[X],HEAD[X]=TOT,FLOW[TOT]=FL; t[++tot]=x,next[tot]=head[y],head[y]=tot,flow[tot]=0;} void Dij () {memset (dis,0x3f,sizeof (dis));d is[1]=0;poq.push (pr (0,1)); int X,y,i;while (!poq.empty ()) {x=poq.top (). Second,poq.pop (); if (Vis[x]) continue;vis[x]=1;for (I=0;i<to[x].size (); i++) {if (!vis[y=to[x][i]]) {if (dis[y]> Dis[x]+val[x][i]) {Dis[y]=dis[x]+val[x][i];p oq.push (pr (dis[y],y));p re[y].clear ();p re[y].push_back (x); else if (Dis[y]==dis[x]+val[x][i]) pre[y].push_back (x);}}} BOOL BFs () {memset (dis,-1,sizeof (dis));d Is[3]=0;q.push (3); int x,y,i;while (!q.empty ()) {X=q.front (), Q.pop (); for (I=head[x];i;i=next[i]) {if (flow[i]>0&&dis[y=t[ i]]<0) {Dis[y]=dis[x]+1;q.push (y);}}} return dis[n<<1]>=0;} ll Dfs (int x,ll k) {if (x== (n<<1)) return k;int y,i;ll j,ret=0;for (I=head[x];i;i=next[i]) {if (flow[i]>0& &dis[y=t[i]]==dis[x]+1) {J=dfs (Y,min (Flow[i],k-ret)); Flow[i]-=j,flow[i^1]+=j;ret+=j;if (K==ret) return ret;}} return ret;} int main () {int m,i,j,k,x,y;ll v,ans;scanf ("%d%d", &n,&m), for (i=1;i<=m;i++) {scanf ("%d%d%lld", &x, &AMP;Y,&AMP;V); To[x].push_back (y), Val[x].push_back (v); To[y].push_back (x), Val[y].push_back (v);} Dij (); Q.push (n); memset (vis,0,sizeof (VIS)), while (!q.empty ()) {X=q.front (), Q.pop (); for (I=0;i<pre[x].size (); i++ {Y=pre[x][i];add (Y&LT;&LT;1|1,X&LT;&LT;1,0X3F3F3F3F3F3F3F3FLL), if (!vis[y]) Vis[y]=1,q.push (y);}} for (i=1;i<=n;i++) scanf ("%lld", &v), add (i<<1,i<<1|1,v), Ans=0;while (BFS ()) Ans+=dfs ( 3,0X3F3F3F3F3F3F3F3FLL);p rintf ("%lld", ans); return 0;}

bzoj-3931 Network throughput