线性规划模板

来源：互联网发布：cfd软件难学吗编辑：程序博客网时间：2024/04/29 14:03
前段时间我参加华为比赛学习的线性规划
虽然现在看来没有卵用
将模板贴一下
#include<cmath>#include<map>#include<iostream>#include<cstring>#include<cstdio>#include<set>#include<vector>#include<algorithm>using namespace std;typedef long long ll;const int MAXN = 1005;const double INF = 1e10;const double eps = 1e-7;int n,m; // 行， 列double c[MAXN]; double b[MAXN]; double a[MAXN][MAXN]; double x[MAXN*2]; double v; int N[MAXN], B[MAXN];int ans;int cnt = 0;void debug(int pos , int flor) {    if(cnt < 10) {        cnt ++;    }else return ;    printf("pos = %d, flor = %d\n",pos,flor);    printf("%d %d %.2f\n",n,m,v);    for(int i = 1; i <= m; ++i)        printf("%.2f ", c[i]); printf("\n");    for(int i = 1; i <= n; ++i) {        printf("%.2f ",b[i]);        for(int j = 1; j <= m; ++j) {            printf("%.2f ",a[i][j]);        }        printf("\n");    }    for(int i = 1; i <= n; ++i) printf("%d ",B[i]); printf("\n");    for(int i = 1; i <= m; ++i) printf("%d ",N[i]); printf("\n");//  for(int i = 1; i <= n+m; ++i) printf("%.2f ",x[i]); printf("\n\n");}int equal(double a, double b) {    double tt = fabs(b-a);    if(tt < eps) return 1;    else return 0;}void priov(int l, int e) { // 第l行，第e列    b[l] /= a[l][e];    for(int i = 1; i <= m; ++i) {        if(i != e) {            a[l][i] /= a[l][e];        }    }        a[l][e] = 1/a[l][e];    for(int i = 1; i <= n; ++i) {        if(i != l && fabs(a[i][e]) > eps) {            b[i] -= b[l] * a[i][e];            for(int j = 1; j <= m; ++j) {                if(j != e) {                    a[i][j] -= a[i][e] * a[l][j];                }            }            a[i][e] = -a[i][e] * a[l][e];        }    }    v += c[e] * b[l];    for(int i = 1; i <= m; ++i) {        if(i != e) {            c[i] -= c[e] * a[l][i];        }    }    c[e] = -c[e] * a[l][e];}double simple(){    while(1) {        int l,e = 0;        for(int i = 1; i <= m; ++i) {            if(c[i] > eps) {                e = i; break;            }           }        if(!e) return v;        double minn = INF;        for(int i = 1; i <= n; ++i) {            if(a[i][e] > eps && minn > b[i]/a[i][e]) {                minn = b[i]/a[i][e]; l = i;             }        }        if(minn == INF) return minn;        priov(l, e);        swap(B[l], N[e]);           }}int initsimplex() { // 找到一个初始可行解    // 备份一些信息    int _m = m; double _v = v;    int _N[m+5]; int _c[m+5];     for(int i = 1; i <= m; ++i) _N[i] = N[i];    for(int i = 1; i <= m; ++i) _c[i] = c[i];    int id, minn = INF;    for(int i = 1; i <= n; ++i)         if(b[i] < minn) {            id = i; minn = b[i];        }    if(b[id] >= 0) return 1;    //最初形式不满足条件，就需要寻找一个    m ++;    v = 0;    for(int i = 1; i <= m; ++i)  c[i] = 0; c[m] = -1;    for(int i = 1; i <= n; ++i) a[i][m] = -1;    N[m] = 0;//  debug(2, 1);    printf("%d %d\n",id,m);    priov(id, m);    swap(B[id], N[m]);    debug(3, 1);    simple();    debug(4,1);    if( equal(v, 0) == 0) return 0;    //现将生成的新松弛型中的x0项删去    for(int i = 1; i <= n; ++i) {        if(B[i] == 0) {            int pos;            for(int j = 1; j <= m; ++j) {                if(a[i][j] != 0) {                    pos = j;                }            }            priov(i, pos);            swap(B[i], N[pos]);            break;        }    }    for(int i = 1; i <= n; ++i) {        int fl = 0;        for(int j = 1; j <= m; ++j) {            if(N[j] == 0) fl = 1;            if(fl) {                a[i][j] = a[i][j+1];             }        }    }    int fl = 0;    for(int i = 1; i <= m; ++i) {        if(N[i] == 0) fl = 1;        if(fl) N[i] = N[i+1];    }    m --;    map<int, int> m1; m1.clear();    map<int, int> m2; m2.clear();    for(int i = 1; i <= n; ++i) m1[B[i]] = i;    for(int i = 1; i <= m; ++i) m2[N[i]] = i;    for(int i = 1; i <= m; ++i) c[i] = 0;    v = _v;    printf("%.3f\n",v);    for(int i = 1; i <= _m; ++i) {        if(m1.count(_N[i])) {            int tt = m1[_N[i]];            for(int j = 1; j <= m; ++j) {                c[j] -= _c[i] * a[tt][j];            }            v += _c[i] * b[tt];        }else {            c[m2[_N[i]]] += _c[i];        }     }    return 1;}void solve(){    while(~scanf("%d %d",&n,&m)) {        ans = -1;        scanf("%lf",&v);        for(int i = 1; i <= m; ++i) scanf("%lf",&c[i]);        for(int i = 1; i <= n; ++i) {            scanf("%lf",&b[i]);            for(int j = 1; j <= m; ++j) {                scanf("%lf",&a[i][j]);            }        }        for(int i = 1; i <= m; ++i) N[i] = i;        for(int i = 1; i <= n; ++i) B[i] = i+m;        set<int> st;    //  for(int i = 1; i <= m; ++i) st.insert(N[i]);        int tt = initsimplex();        for(int i = 1; i <= m; ++i) st.insert(N[i]);        if(!tt) {            printf("there is none true answer!\n");            return;        }        debug(1,1);        simple();        memset(x, 0, sizeof(x));        for(int i = 1; i <= n; ++i) {            if(st.find(B[i]) != st.end()) {                x[B[i]] = b[i];            }        }        debug(1, 1);        printf("ans = %d \n", ans);    }}int main(){    solve();    return 0;}/*   2 2    0   3 13   40 2 9   82 11 -8   5 2 0.00   3.00 13.00    40.00 2.00 9.00    82.00 11.00 -8.00    9.00 1.00 0.00    2.00 0.00 1.00    -9.00 -1.00 0.00   4 2 0.00   3.00 13.00    40.00 2.00 9.00    82.00 11.00 -8.00    9.00 1.00 0.00    -3.00 0.00 -1.00 */
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