Bzoj 3112: [Zjoi2013] Defensive front [simplex method]

Title Description

The front line can be regarded as a sequence of length n, now it is necessary to build towers on this sequence to defend the enemy soldiers, to build a tower with CI in the Sequence No. I position, and a position can be built any number of towers, cost cumulative calculation. There are m zones [L1, R1], [L2, R2], ..., [Lm, Rm], at least di tower in the range of section I. Minimum cost.

Input/output format

Input format:

The first behavior is two numbers n, M.

On the next line, there are n numbers, which describe the C array.

Next m line, three numbers per line Li,ri,di, describes an interval.

Output format:

Contains only one row, a number, and a minimum cost.

Input and Output Sample input example # #:

5 31 5 6 3 42 3 11 5 43 5 2

Sample # # of output:

11

Description

"Sample description"

Location 1 built 2 towers, location 3 built a tower, location 4 built a tower. Cost 1*2+6+3=11.

"Data Size"

For 20% of data, n≤20,m≤20.

For 50% of the data (including the last part of the data), Di is all 1.

For 70% of the data (including the previous part of the data), n≤100,m≤1000.

For 100% of the data, N≤1000,m≤10000,1≤li≤ri≤n, the rest of the data are ≤10000.

It's like 08 volunteer recruiting.

Set X as the vector of whether the tower is built for each location, 1 build 0 not built

Minimize CX

Satisfy constraint I constraint is x[l[i]]+x[l[i]+1]+...+x[r[i]]>=d[i]

namely Ax>=d

A[I][J] for 1 means that the first constraint j in [L[i],r[i]], can contribute to the

After the duality, it becomes

Maximize DX

Meet constrained at X<=c

Note that this time is n constraints, m variables

////main.cpp//zjoi2013 Defensive Front////Created by Candy on 2016/12/16.//copyright©2016 year Candy. All rights reserved.//#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespaceStd;typedefLong Longll;Const intm=10005, n=1005, inf=1e9;Const Doubleeps=1e-6; inlineintRead () {CharC=getchar ();intx=0, f=1;  while(c<'0'|| C>'9'){if(c=='-') f=-1; C=GetChar ();}  while(c>='0'&&c<='9') {x=x*Ten+c-'0'; C=GetChar ();} returnx*F;}intn,m;Doublea[n][m],b[n],c[m],v;voidPivotintLinte) {B[l]/=A[l][e];  for(intI=1; i<=n;i++)if(i!=e) a[l][i]/=A[l][e]; A[l][e]=A[l][e];  for(intI=1; i<=m;i++)if(I!=l&&fabs (a[i][e]) >0) {B[i]-=a[i][e]*B[l];  for(intj=1; j<=n;j++)if(j!=e) a[i][j]-=a[i][e]*A[l][j]; A[i][e]=-a[i][e]*A[l][e]; } v+=c[e]*B[l];  for(intj=1; j<=n;j++)if(j!=e) c[j]-=c[e]*A[l][j]; C[e]=-c[e]*a[l][e];}voidsimplex () { while(true){        intE=0, l=0; Doublemn=INF;  for(e=1; e<=n;e++)if(c[e]>eps) Break; if(e==n+1)return;  for(intI=1; i<=m;i++)            if(A[i][e]>eps&&mn>b[i]/a[i][e]) mn=b[i]/a[i][e],l=i; if(Mn==inf)return;//unboundedpivot (l,e); }}intMainintargcConst Char*argv[]) {N=read (); m=read ();  for(intI=1; i<=n;i++) scanf ("%LF",&B[i]);  for(intI=1; i<=m;i++){        intL=read (), r=read ();  for(intj=l;j<=r;j++) a[j][i]=1.0; scanf ("%LF",&C[i]);    } swap (M,n);    Simplex (); printf ("%d",(int) (v+0.5)); return 0;}

Bzoj 3112: [Zjoi2013] Defensive front [simplex method]