Submit: 5725 Solved: 3437
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Description
Input
Output
仅包含一个整数,表示你所设计的最优方案的总费用。
Sample Input
2 3 4
1 2 2
2 3 5
3 3 2
Sample Output
HINT
1 ≤ N ≤ 1000,1 ≤ M ≤ 10000,题目中其他所涉及的数据均 不超过2^31-1。
Source
如果不知道这题是线性规划的话肯定很难看出来,不过知道了就好做多了
若$C_i$为第$i$个人的花费,$a_i$为第$i$天需要的人,$x_i$为第$i$个人的数量
那么我们需要满足对于每一天$i$,$\sum_{i = 1}^{M} x_i >= a_i$,同时$\sum C_i x_i$最小
啥?最小?当时我推出式子来就蒙了qwq。然后跑去膜题解
根据对偶原理,问题相当于使得$\sum_{i = 1}^{M} x_i <= C_i$,的情况下$\sum a_i x_i$最大
仔细一想好像挺有道理
关于最后答案是否为整数的问题
https://www.luogu.org/problemnew/solution/P3980
#include<cstdio> #include<algorithm> #include<cmath> #define LL long long using namespace std; const int MAXN = 51, INF = 1e9 + 10; const double eps = 1e-8; inline int read() { char c = getchar();int x = 0,f = 1; while(c < '0' || c > '9'){if(c == '-')f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = x * 10 + c - '0',c = getchar();} return x * f; } int N, M; LL a[10001][1001]; void Pivot(int l, int e) { double t = a[l][e]; a[l][e] = 1; for(int i = 0; i <= N; i++) a[l][i] /= t; for(int i = 0; i <= M; i++) { if(i != l && abs(a[i][e]) > eps) { t = a[i][e]; a[i][e] = 0; for(int j = 0; j <= N; j++) a[i][j] -= a[l][j] * t; } } } bool simplex() { while(1) { int l = 0, e = 0; double mn = INF; for(int i = 1; i <= N; i++) if(a[0][i] > eps) {e = i; break;} if(!e) break; for(int i = 1; i <= M; i++) if(a[i][e] > eps && a[i][0] / a[i][e] < mn) mn = a[i][0] / a[i][e], l = i; Pivot(l, e); } return 1; } int main() { srand(19260817); N = read(); M = read(); for(int i = 1; i <= N; i++) a[0][i] = read(); for(int i = 1; i <= M; i++) { int S = read(), T = read(), C = read(); for(int j = S; j <= T; j++) a[i][j] = 1; a[i][0] = C; } simplex(); printf("%lld", -a[0][0]); return 0; }