# The "BZOJ3876" "Ahoi2014" spur plot has the lowest cost maximum flow in the Nether

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Author: User

`#include <stdio.h>int main () {puts ("reprint please specify source thank you");p UTS ("http://blog.csdn.net/vmurder/article/details/43025375 ");}`

"BZOJ2324" rescue Pikachu This problem is also a lower-bound minimum cost maximum flow.

This problem is actually a template problem.

My approach is to take each side of a flow of 1 out, and then the cost of the original cost-inf. In the construction of the diagram can be deducted from these INF added back. It can be proved that this method will not be stuck in a negative loop at least on the topology map.

Code:

`#include <queue> #include <cstdio> #include <cstring> #include <iostream> #include < algorithm> #define N 500#define M 31000#define inf 0x3f3f3f3f#define inf 0x3f3f3f3f3f3f3f3fllusing namespace std; struct ksd{int u,v,len,next;long long fee;} E[m];int head[n],cnt;void Add (int u,int v,int len,int fee) {Cnt++;e[cnt].u=u;e[cnt].v=v;e[cnt].len=len;e[cnt].fee=fee ; e[cnt].next=head[u];head[u]=cnt;} int S,t;long long dist[n];int pre[n],lim[n];bool in[n];queue<int>q;void SPFA () {memset (dist,0x3f,sizeof (Dist)); memset (Lim,0,sizeof (LIM)), while (!q.empty ()) Q.pop (), int i,u,v;dist[s]=0,in[s]=1;lim[s]=inf;q.push (s), while (! Q.empty ()) {U=q.front (), Q.pop (), In[u]=0;for (I=head[u];i;i=e[i].next) {v=e[i].v;if (!e[i].len) continue;if (Dist[v] >dist[u]+e[i].fee) {dist[v]=dist[u]+e[i].fee;lim[v]=min (E[i].len,lim[u]);p re[v]=i;if (!in[v]) {In[v]=1;q.push ( v);}}} return;} void handle (int flow) {for (int i=pre[t];i;i=pre[e[i].u]) {e[i].len-=flow;e[i^1].len+=flow;}} Long Long minfee;int n,m;void build (){int I,j,k;int a,b,c;cnt=1;scanf ("%d", &n), S=n+1,t=n+2;add (s,1,inf,0), add (1,s,0,0); for (i=1;i<=n;i++) {scanf ("%d", &a), add (i,t,inf,0), add (t,i,0,0), while (a--) {scanf ("%d%d", &b,&c); Minfee+=inf;add (I,b,inf,c), Add (b,i,0,-c); Add (I,b,1,c-inf), add (B,i,0,inf-c);}} return;} void Checkedge () {for (int u=1;u<=t;u++) {printf (' from%d: ', u), for (int i=head[u];i;i=e[i].next) {printf ("%d/%d/%d", E[i].v,e[i].len,e[i].fee);} Puts ("");} Puts ("");} int main () {freopen ("test.in", "R", stdin); build (); while (SPFA (), Dist[t]<inf) {Minfee+=dist[t]*lim[t];handle (lim[ T]);} Cout<<minfee<<endl;return 0;}`

The "BZOJ3876" "Ahoi2014" spur plot has the lowest cost maximum flow in the Nether

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