【BZOJ1061】[Noi2008]志願者招募【單純形法】
阿新 • • 發佈:2019-01-03
雙倍經驗題,BZOJ3265。
先用對偶原則轉換成求對偶問題的解,這樣直接轉化成了標準型,然後跑Simplex就好了。
下面是對樣例的一個計算過程。
/* Footprints In The Blood Soaked Snow */ #include <cstdio> #include <cmath> typedef double DB; typedef long long LL; const int maxn = 1005, maxm = 10005; const DB inf = 0x3f3f3f3f3f3f3f3f, eps = 1e-7; int n, m; DB b[maxm], c[maxn], cof[maxm][maxn], ans; inline void pivot(int id, int pos) { b[id] /= cof[id][pos]; cof[id][pos] = 1 / cof[id][pos]; for(int i = 1; i <= n; i++) if(i != pos) cof[id][i] *= cof[id][pos]; for(int i = 1; i <= m; i++) if(i != id && fabs(cof[i][pos]) > eps) { b[i] -= cof[i][pos] * b[id]; for(int j = 1; j <= n; j++) if(j != pos) cof[i][j] -= cof[i][pos] * cof[id][j]; cof[i][pos] = -cof[i][pos] * cof[id][pos]; } ans += c[pos] * b[id]; for(int i = 1; i <= n; i++) if(i != pos) c[i] -= c[pos] * cof[id][i]; c[pos] = -c[pos] * cof[id][pos]; } inline DB simplex() { while(1) { int pos, id; for(pos = 1; pos <= n; pos++) if(c[pos] > eps) break; if(pos == n + 1) return ans; DB tmp = inf; for(int i = 1; i <= m; i++) if(cof[i][pos] > eps && b[i] / cof[i][pos] < tmp) tmp = b[i] / cof[i][pos], id = i; if(tmp == inf) return inf; pivot(id, pos); } } int main() { scanf("%d%d", &n, &m); for(int i = 1; i <= n; i++) scanf("%lf", &c[i]); for(int i = 1; i <= m; i++) { int x, y; scanf("%d%d", &x, &y); for(int j = x; j <= y; j++) cof[i][j] = 1; scanf("%lf", &b[i]); } printf("%lld\n", LL(simplex() + 0.5)); return 0; }