"BZOJ1061" "NOI2008" volunteer recruitment cost flow divine question, simplex nude question (code cost flow)

The problem model out of the true God.

It needs to be very difficult to roll out.

Originally intended to write a good problem, but the state is not good, did not get this question.

You can only leave a hole in the building.

It is said that the "simplex algorithm" can be high-speed + bare-figure water over this problem (er, or the problem is a simplex naked question is also. ）

Stay Pits before you give a link, it should be the best problem on the Internet now:

the byv of the Great God:www.byvoid.com/blog/noi-2008-employee/#more-916

My Code:

<span style= "FONT-FAMILY:KAITI_GB2312;FONT-SIZE:18PX;" > #include <queue> #include <cstdio> #include <cstring> #include <iostream> #include < algorithm> #define N 1050#define M 30100#define P 105#define inf 0x3f3f3f3fusing namespace std;struct ksd{int U,v,len,fe E,next;} 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,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;}} int N,m,minfee; int Need[n];int Main () {//freopen ("test.in", "R", stdin), int i,a,b,c;scanf ("%d%d", &n,&m), n++;s=n+1,t=n+2, Cnt=1;for (i=1;i<n;i++) scanf ("%d", &need[i]); for (i=1;i<=m;i++) {scanf ("%d%d%d", &a,&b,&c); Add (A,b+1,inf,c), add (b+1,a,0,-c);} for (i=1;i<=n;i++) {c=need[i]-need[i-1];if (c>=0) Add (s,i,c,0), add (i,s,0,0), else Add (i,t,-c,0), add (t,i,0,0); if (i>1) Add (i,i-1,inf,0), add (i-1,i,0,0);} while (SPFA (), Dist[t]<inf) {minfee+=dist[t]*lim[t];handle (lim[t]);} printf ("%d\n", Minfee); return 0;} </span>

"BZOJ1061" "NOI2008" volunteer recruitment cost flow divine question, simplex nude question (code cost flow)