【BZOJ1061】[Noi2008]志愿者招募【单纯形法】

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双倍经验题，BZOJ3265。

先用对偶原则转换成求对偶问题的解，这样直接转化成了标准型，然后跑Simplex就好了。

下面是对样例的一个计算过程。

/* 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;}

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