Bzoj 3996 [Tjoi 2015] linear algebra problem Solving report

First, we can get:

$ $D = \sum_{i=1}^{n}\sum_{j=1}^{n}a_i\times a_j\times b_{i,j}-\sum_{i=1}^{n}a_i\times c_i$$

So is it the equivalent of the question:

There are $n $ items, you can choose some items to come out, if you chose $i at the same time, j$ two items then there is $b _{i,j}$ benefits, but every item has a price $c _i$, to seek the maximum benefit.

This is the classic minimal cut model:

• The Edge $S \to Dot (i,j) $ with a flow rate of $b _{i,j}$.
• The Edge $Dot (i,j) \to i$ and the $Dot (i,j) \to j$ with a traffic of $\infty$.
• Edge $i \to t$, flow rate is $c _i$

Set the minimum cut to $x $, then the answer is:

$$\SUM_{I=1}^{N}\SUM_{J=1}^{N}B_{I,J}-x$$

Although there are a lot of points, but can still run fast. (*^_^*)

1#include <cmath>2#include <cstdio>3#include <cstring>4#include <iostream>5#include <algorithm>6 using namespacestd;7typedefLong LongLL;8 #defineN 250000 + 500 + 59 #defineM 2500000 + 5Ten #defineINF 0x7fffffff One  A intN, S, T, cnt, tot =1; - intHead[n], q[n], dfn[n]; - LL ans; the  - structEdge - { -     intNext, node, flow; + }h[m]; -  +InlinevoidAddedge (intUintVintFL) A { ath[++ Tot].next = Head[u], head[u] =tot; -H[tot].node = V, H[tot].flow =FL; -h[++ Tot].next = Head[v], head[v] =tot; -H[tot].node = u, H[tot].flow =0; - } -  inInlineBOOLBFS () - { to      for(inti = S; I <= T; i + +) +Dfn[i] =0; -     intL =1, r =1; theq[1] = S, dfn[s] =1; *      while(L <=R) $     {Panax Notoginseng         intz = q[l + +]; -          for(inti = head[z]; I i =h[i].next) the         { +             intd = h[i].node, p =H[i].flow; A             if(!p | | DFN[D])Continue ; theDFN[D] = Dfn[z] +1; +q[++ R] =D; -             if(d = = T)return 1; $         } $     } -     return 0; - } the  -InlineintDinic (intZintinflow)Wuyi { the     if(z = = T)returninflow; -     intRET =inflow, flow; Wu      for(inti = head[z]; I i =h[i].next) -     { About         intd = h[i].node, p =H[i].flow; $         if(Dfn[d]! = Dfn[z] +1|| !P)Continue ; -Flow =Dinic (d, Min (P, ret)); -RET-=flow; -H[i].flow-= flow, H[i ^1].flow + =flow; A         if(!ret)returninflow; +     } the     if(ret = = inflow) DFN[Z] =-1; -     returnInflow-ret; $ } the  the intMain () the { the #ifndef Online_judge -Freopen ("3996.in","R", stdin); inFreopen ("3996.out","W", stdout); the     #endif the      Aboutscanf"%d", &n); theS =0, T = n + N * n +1, cnt =N; the      for(inti =1; I <= N; i + +) the          for(intj =1, W; J <= N; J + +) +         { -scanf"%d", &W); theAddedge (S, + +)CNT, W);Bayi Addedge (CNT, I, INF); the Addedge (CNT, J, INF); theAns + =W; -         } -      for(inti =1, W; I <= N; i + +) the     { thescanf"%d", &W); the Addedge (i, T, W); the     } -      while(BFS ()) theAns-=dinic (S, INF); theprintf"%lld\n", ans); the     94 #ifndef Online_judge the fclose (stdin); the fclose (stdout); the     #endif98     return 0; About}

3996_gromah

Bzoj 3996 [Tjoi 2015] linear algebra problem Solving report