我是靠谱客的博主 聪明翅膀,这篇文章主要介绍单纯形模板,现在分享给大家,希望可以做个参考。

uoj179
输入后转化为线性规划标准形式
[max ; z = sumlimits_{j = 1}^{n}c_jx_j]
[ left{ begin{aligned} sumlimits_{j = 1}^{n}a_{ij}x_j = b_j quad i in [1,m] \ x_j ge 0 quad j in [1,n] end{aligned} right. ]
其中初始化操作当且仅当初始解({0,0,dots,0})不合法时调用

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#include<algorithm> #include<iostream> #include<cstring> #include<cstdlib> #include<cstdio> #include<ctime> #include<cmath> #include<map> #define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt) #define REP(i,n) for (int i = 1; i <= (n); i++) #define mp(a,b) make_pair<int,int>(a,b) #define cls(s) memset(s,0,sizeof(s)) #define cp pair<int,int> #define LL long long int using namespace std; const int maxn = 25,maxm = 100005; const double eps = 1e-8,INF = 1e15; inline int read(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();} return out * flag; } int n,m,id[maxn << 1]; double a[maxn][maxn],ans[maxn]; void Pivot(int l,int e){ swap(id[n + l],id[e]); double t = a[l][e]; a[l][e] = 1; for (int j = 0; j <= n; j++) a[l][j] /= 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 init(){ while (true){ int e = 0,l = 0; for (int i = 1; i <= m; i++) if (a[i][0] < -eps && (!l|| (rand() & 1))) l = i; if (!l) break; for (int j = 1; j <= n; j++) if (a[l][j] < -eps && (!e || (rand() & 1))) e = j; if (!e){puts("Infeasible"); return false;} Pivot(l,e); } return true; } bool simplex(){ while (true){ int l = 0,e = 0; double mn = INF; for (int j = 1; j <= n; j++) if (a[0][j] > eps){e = j; 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; if (!l){puts("Unbounded"); return false;} Pivot(l,e); } return true; } int main(){ srand(time(NULL)); n = read(); m = read(); int t = read(); REP(i,n) a[0][i] = read(); REP(i,m){ REP(j,n) a[i][j] = read(); a[i][0] = read(); } REP(i,n) id[i] = i; if (init() && simplex()){ printf("%.8lfn",-a[0][0]); if (t){ REP(i,m) ans[id[n + i]] = a[i][0]; REP(i,n) printf("%.8lf ",ans[i]); } } return 0; }

转载于:https://www.cnblogs.com/Mychael/p/9247942.html

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