bzoj3112 [Zjoi2013]防守战线

Posted 2020-09-01 wfj_2048

正解：线性规划。

直接套单纯形的板子，因为所约束条件都是>=号，且目标函数为最小值，所以考虑对偶转换，转置一下原矩阵就好了。

 

 1 //It is made by wfj_2048~
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <complex>
 5 #include <cstring>
 6 #include <cstdlib>
 7 #include <cstdio>
 8 #include <vector>
 9 #include <cmath>
10 #include <queue>
11 #include <stack>
12 #include <map>
13 #include <set>
14 #define inf (1e15)
15 #define eps (1e-12)
16 #define il inline
17 #define RG register
18 #define ll long long
19 
20 using namespace std;
21 
22 double a[1010][10010];
23 int b[10010],n,m;
24 
25 il int gi(){
26     RG int x=0,q=1; RG char ch=getchar(); while ((ch<‘0‘ || ch>‘9‘) && ch!=‘-‘) ch=getchar();
27     if (ch==‘-‘) q=-1,ch=getchar(); while (ch>=‘0‘ && ch<=‘9‘) x=x*10+ch-48,ch=getchar(); return q*x;
28 }
29 
30 il void pivot(RG int l,RG int e){
31     RG double k=a[l][e]; a[l][e]=1;
32     for (RG int i=0;i<=n;++i) a[l][i]/=k; RG int len=0;
33     for (RG int i=0;i<=n;++i) if (fabs(a[l][i])>eps) b[++len]=i;
34     for (RG int i=0;i<=m;++i)
35     if (i!=l && fabs(a[i][e])>eps){
36         k=a[i][e],a[i][e]=0;
37         for (RG int j=1;j<=len;++j) a[i][b[j]]-=k*a[l][b[j]];
38     }
39     return;
40 }
41 
42 il double simplex(){
43     while (1){
44     RG int l,e; for (e=1;e<=n;++e) if (a[0][e]>eps) break;
45     if (e==n+1) return -a[0][0]; RG double tmp=inf;
46     for (RG int i=1;i<=m;++i)
47         if (a[i][e]>eps && tmp>a[i][0]/a[i][e]) tmp=a[i][0]/a[i][e],l=i;
48     if (tmp==inf) return inf; pivot(l,e);
49     }
50 }
51 
52 il void work(){
53     m=gi(),n=gi(); RG int l,r,d;
54     for (RG int i=1;i<=m;++i) a[i][0]=gi();
55     for (RG int i=1;i<=n;++i){
56     l=gi(),r=gi(),d=gi(); a[0][i]=d;
57     for (RG int j=l;j<=r;++j) a[j][i]=1;
58     }
59     printf("%lld\n",(ll)(simplex()+0.5)); return;
60 }
61 
62 int main(){
63     work();
64     return 0;
65 }