BZOJ1003 Logistics Transportation

Test instructions: There are n days, M points, every day to go from 1 to M, some points can not be reached at some time, to ensure that every day there must be a road to M, each change route to spend K, to find the minimum cost of how much

This problem data range is very small, random mess.

Run a n^2 SPFA first, find out cost[i][j] (the minimum cost per day of the day to the first J Day)

One more DP will be done:

F[i]=min (f[i],f[j]+cost[j+1][i]* (i-j) +k);

1#include <cstdio>2#include <cstring>3#include <iostream>4#include <queue>5 using namespacestd;6 structEdge {7     intv,next,w;8}e[2005];9 intk=1, N,m,e,d,kk;Ten inttim[ -][ the],avl[ -],dis[ -],vis[ the],head[2005],f[ the],cost[ the][ the]; One voidAdde (intUintVintW) { Ae[k].v=v;e[k].w=w;e[k].next=head[u];head[u]=k++; - } -Queue <int>Q; the intSPFA (intXinty) { -memset (DIS, the,sizeof(DIS)); -memset (AVL,0,sizeof(AVL)); -memset (Vis,0,sizeof(Vis)); +      for(intI=1; i<=m;i++) -          for(intj=x;j<=y;j++) +             if(Tim[i][j]) avl[i]=1; AQ.push (1); vis[1]=1;d is[1]=0; at      while(!Q.empty ()) { -         intU=q.front (); Q.pop (); vis[u]=0; -          for(intI=head[u];~i;i=E[i].next) { -             intv=e[i].v; -             if(!avl[v]&&dis[v]>dis[u]+E[I].W) { -dis[v]=dis[u]+E[I].W; in                 if(!Vis[v]) { -Q.push (v); vis[v]=1; to                 } +             } -         } the     } *     returnDis[m]; $ }Panax Notoginseng  - intMain () the { +memset (head,-1,sizeof(head)); AMemset (F, the,sizeof(f)); thef[0]=0; +scanf"%d%d%d%d",&n,&m,&kk,&E); -      for(intI=1, a,b,c;i<=e;i++) { $scanf"%d%d%d",&a,&b,&c); $ Adde (a,b,c); Adde (b,a,c); -     } -scanf"%d",&d); the      for(intI=1, p,a,b;i<=d;i++) { -scanf"%d%d%d",&p,&a,&b);Wuyi          for(intj=a;j<=b;j++) tim[p][j]=1; the     } -      for(intI=1; i<=n;i++) Wu          for(intj=1; j<=n;j++) -cost[i][j]=SPFA (i,j); About      for(intI=1; i<=n;i++) $          for(intj=0; j<i;j++) -             if(cost[j+1][i]!=0x3f3f3f3f) -F[i]=min (f[i],f[j]+cost[j+1][i]* (I-J) +KK); -printf"%d\n", f[n]-KK); A     return 0; +}

View Code

BZOJ1003 Logistics Transportation