• BZOJ1061 [Noi2008]志愿者招募 【单纯形】


    题目链接

    BZOJ1061

    题解

    今天终于用正宗的线性规划(A)了这道题
    题目可以看做有(N)个限制和(M)个变量
    变量(x_i)表示第(i)种志愿者的人数,对于第(i)种志愿者所能触及的那些天,(x_i)的系数都为(1),其余为(0)
    也就是

    [min ; z = sumlimits_{i = 1}^{M} C_ix_i \ left{ egin{aligned} sumlimits_{i = 1}^{M} [S_i le j le T_i]x_i ge A_i qquad j in [1,N]\ x_i ge 0 qquad i in [1,M] end{aligned} ight. ]

    转化为标准型线性规划,使用单纯形算法求解即可
    诶?解保证是整数吗?

    似乎相对于费用流,空间大且跑得慢,,,

    #include<algorithm>
    #include<iostream>
    #include<cstring>
    #include<cstdlib>
    #include<cstdio>
    #include<vector>
    #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 N = 1005,M = 10005;
    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[M << 1];
    double a[N][M];
    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 && 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)); int S,T,C;
    	m = read(); n = read();
    	REP(i,m) a[i][0] = -read();
    	REP(j,n){
    		S = read(); T = read(); C = read();
    		for (int i = S; i <= T; i++)
    			a[i][j] = -1;
    		a[0][j] = -C;
    	}
    	REP(i,n) id[i] = i;
    	init(); simplex();
    	printf("%d",(int)(a[0][0] + 0.5));
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/Mychael/p/9248381.html
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