BZOJ3265: 志愿者招募加强版(线性规划)

最新推荐文章于 2019-06-29 18:56:19 发布

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最新推荐文章于 2019-06-29 18:56:19 发布

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原文链接：http://www.cnblogs.com/zwfymqz/p/9320584.html

本文探讨了使用线性规划解决复杂问题的高效方法，通过具体实例解析了如何利用简单的循环改进算法性能，并深入介绍了线性规划求解过程中的关键步骤，如基变量的选择和迭代更新。

Time Limit: 20 Sec Memory Limit: 512 MB
Submit: 809 Solved: 417
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Description

Input

Output

Sample Input

3 3
2 3 4
1 1 2 2
1 2 3 5
1 3 3 2

Sample Output

HINT

Source

这题是来搞笑的么？？

除了多循环几次之外和原版有啥区别？qwq、

#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() {
//    freopen("a.in", "r", stdin);
    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 K = read();
        while(K--) {
            int S = read(), T = read();
            for(int j = S; j <= T; j++)    
                a[i][j] = 1;        
        }
        int C = read();
        a[i][0] = C;
    }
    simplex();
    printf("%lld", -a[0][0]);
    return 0;
}