BZOJ3550 [ONTAK2010]Vacation 【单纯形】

题目链接

BZOJ3550

题解

单纯形裸题

题意不清，每个位置最多选一次

#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 = 1005,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,N,K;

double a[maxn][maxn];

void Pivot(int l,int 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 && fabs(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;

	}

}

void 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;

		Pivot(l,e);

	}

}

void 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;

		Pivot(l,e);

	}

}

int main(){

	srand(time(NULL));

	N = read(); n = 3 * N; K = read();

	int tmp = (N << 1 | 1); m = tmp + n;

	for (int j = 1; j <= n; j++){

		a[0][j] = read();

		a[tmp + j][j] = 1;

		a[tmp + j][0] = 1;

	}

	for (int j = 1; j <= n; j++){

		int E = min(j,tmp);

		for (int i = max(1,j - N + 1); i <= E; i++)

			a[i][j] = 1;

	}

	for (int i = 1; i <= tmp; i++) a[i][0] = K;

	init(); simplex();

	printf("%lld\n",(LL)(-a[0][0] + 0.5));

	return 0;

}