单纯性与网络流

单纯性模板：

需要b > 0, xi > 0.

 1 // 单纯性
 2 // n+1 * m+1 矩阵
 3 // 1~n 为约束 <= 右值
 4 // n+1为目标最大值
 5 // 对偶化直接转置即可
 6 const int maxn = 1100, maxm = 11000;
 7 const double eps = 1e-7, INF = 1e20;
 8 int dcmp(double a) { return fabs(a)<eps?0:a<0?-1:1; }
 9 int n , m;
10 double a[maxm][maxn];
11 void pivot(int l , int e){
12     for(int i = 1; i <= n; i++) if(i != l&&dcmp(a[i][e])){
13         for(int j = 1; j <= m; j++)
14             if(j != e) a[i][j] -= a[i][e]/a[l][e]*a[l][j];
15         a[i][e] /= -a[l][e];
16     }
17     for(int i = 1; i <= m; i++) if(i != e) a[l][i] /= a[l][e]; a[l][e] = 1/a[l][e];
18 }
19
20 double simplex(){
21     double mn;
22     int e , l;
23     while(1){
24         for(e=1;e<m;e++) if(dcmp(a[n][e]) > 0) break;
25         if(e == m) return -a[n][m];
26         mn = INF;
27         for(int i=1;i<n;i++) if(dcmp(a[i][e]) > 0 && mn > a[i][m]/a[i][e]) mn = a[l=i][m]/a[i][e];
28         if(mn == INF) return INF;
29         pivot(l, e);
30     }
31 }

bzoj1061

题意：有n天，第i天需要ai个志愿者，有m类志愿者，每类志愿者工作时间为[l,r]，花费为ci，求最小花费

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3
 4 const int maxn = 1100, maxm = 11000;
 5 const double eps = 1e-7, INF = 1e20;
 6 int dcmp(double a) { return fabs(a)<eps?0:a<0?-1:1; }
 7 int n , m;
 8 double a[maxm][maxn];
 9 void pivot(int l , int e){
10     for(int i = 1; i <= n; i++) if(i != l&&dcmp(a[i][e])){
11         for(int j = 1; j <= m; j++)
12             if(j != e) a[i][j] -= a[i][e]/a[l][e]*a[l][j];
13         a[i][e] /= -a[l][e];
14     }
15     for(int i = 1; i <= m; i++) if(i != e) a[l][i] /= a[l][e]; a[l][e] = 1/a[l][e];
16 }
17
18 double simplex(){
19     double mn;
20     int e, l;
21     while(1){
22         for(e=1;e<m;e++) if(dcmp(a[n][e]) > 0) break;
23         if(e == m) return -a[n][m];
24         mn = INF;
25         for(int i=1;i<n;i++) if(dcmp(a[i][e]) > 0 && mn > a[i][m]/a[i][e]) mn = a[l=i][m]/a[i][e];
26         if(mn == INF) return INF;
27         pivot(l, e);
28     }
29 }
30
31 int main() {
32     scanf("%d%d", &n, &m);
33     //对偶化
34     for(int i = 1; i <= n; i++)
35         scanf("%lf", &a[m+1][i]);
36     for(int i = 1, l, r, v; i <= m; i++) {
37         scanf("%d%d%d" , &l, &r, &v);
38         for(int j = l; j <= r; j++) a[i][j] = 1;
39         a[i][n+1] = v;
40     }
41     n++, m++;
42     swap(n, m);
43     printf("%d" , (int)(simplex()+0.50));
44     return 0;
45 }