BZOJ 3112([Zjoi2013]防守战线-单纯形)

战线可以看作一个长度为n 的序列，现在需要在这个序列上建塔来防守敌兵，在序列第i 号位置上建一座塔有Ci 的花费，且一个位置可以建任意多的塔，费用累加计算。有m 个区间[L1, R1], [L2, R2], …, [Lm, Rm]，在第i 个区间的范围内要建至少Di 座塔。求最少花费。

简单的模板题

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} int m,n;#define EPS (1E-7)#define MAXM (1000+10)#define MAXN (10000+10)namespace Linear_Programming{      double A[MAXM][MAXN],b[MAXM],c[MAXN],v;      void Pivot(int l,int e)      {          int i,j;          b[l]/=A[l][e];          for(i=1;i<=n;i++)              if(i!=e)                  A[l][i]/=A[l][e];          A[l][e]=1/A[l][e];          for(i=1;i<=m;i++)              if(i!=l&&fabs(A[i][e])>EPS)              {                  b[i]-=A[i][e]*b[l];                  for(j=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(i=1;i<=n;i++)              if(i!=e)                  c[i]-=c[e]*A[l][i];          c[e]=-c[e]*A[l][e];      }      double Simplex()      {          int i,l,e;          while(1)          {              for(i=1;i<=n;i++)                  if(c[i]>EPS)                      break;              if((e=i)==n+1)                  return v;              double temp=INF;              for(i=1;i<=m;i++)                  if( A[i][e]>EPS && b[i]/A[i][e]<temp )                      temp=b[i]/A[i][e],l=i;              if(temp==INF) return INF;              Pivot(l,e);          }      }  }  /*input: n=±äÔªÊý m=·½³ÌÊýAx<=b x>=0 Çómin(Cx)Ax>=b x>=0 Çómax(Cx)?AתÖã¬b,c½»»»  n,m½»»»È»ºóÕÕ×ö¾ÍÐÐ */int main(){//  freopen("bzoj3112.in","r",stdin);//  freopen(".out","w",stdout);    using namespace Linear_Programming;    m=read(),n=read();    For(i,m) {          scanf("%lf",&b[i]);    }       For(i,n) {        int x=read(),y=read(),z=read();        for(int j=x;j<=y;j++) A[j][i]=1;        c[i]=z;    }     double ans=Simplex();      printf("%d\n",int(ans+0.5));      return 0;}