题意：一个矩阵，已知其中一些格会降落伞兵，每行每列都有一个武器，可以一次性消灭该行或该列的全部伞兵，每个武器对应不同的价格，若使用多个武器则总价是各个武器价钱的乘积，问消灭所有伞兵最少要多少钱。
分析：最小权覆盖集。难点在于武器总价不是加和而是乘积，那么我们需要把各个单价转化为以e为底的对数，这样再求对数加和的时候其实各个原单价之间是相乘的关系。建立二分图，每行对应一个节点，每列对应一个节点，各点权值等于其武器价钱的以e为底的对数，每个伞兵是连接其所在行列节点的一条边。对这个二分图求最小权独立集。转化为最小割，转化为最大流。
View Code #include#include #include #include #include using namespace std;#define inf 0x3f3f3f3f#define maxn 55#define maxl 505#define N maxn * 2#define E (N + maxl) * 2int n, m, num;int s, t;struct edge{ int x, y, nxt; double c;}bf[E];int ne, head[N], cur[N], ps[N], dep[N];void addedge1(int x, int y, double c){ bf[ne].x = x; bf[ne].y = y; bf[ne].c = c; bf[ne].nxt = head[x]; head[x] = ne++;}void addedge(int x, int y, double c){ addedge1(x, y, c); addedge1(y, x, 0);}double flow(int n, int s, int t){ double tr, res = 0; int i, j, k, f, r, top; while (1) { memset(dep, -1, n * sizeof(int)); for (f = dep[ps[0] = s] = 0, r = 1; f !=r;) for (i = ps[f++], j = head[i]; j; j = bf[j].nxt) { if (bf[j].c && -1 == dep[k = bf[j].y]) { dep[k] = dep[i] + 1; ps[r++] = k; if (k == t) { f = r; break; } } } if (-1 == dep[t]) break; memcpy(cur, head, n * sizeof(int)); for (i = s, top = 0;;) { if (i == t) { for (k = 0, tr = inf; k < top; ++k) if (bf[ps[k]].c < tr) tr = bf[ps[f = k]].c; for (k = 0; k < top; ++k) bf[ps[k]].c -=tr, bf[ps[k]^1].c += tr; res += tr; i = bf[ps[top = f]].x; } for (j = cur[i]; cur[i]; j = cur[i] = bf[cur[i]].nxt) if (bf[j].c && dep[i] + 1 == dep[bf[j].y]) break; if (cur[i]) { ps[top++] = cur[i]; i = bf[cur[i]].y; } else { if (0 == top) break; dep[i] = -1; i = bf[ps[--top]].x; } } } return res;}void input(){ scanf("%d%d%d", &n, &m, &num); s = 0; t = n + m + 1; for (int i = 1; i <= n; i++) { double a; scanf("%lf", &a); addedge(s, i, log(a)); } for (int i = n + 1; i <= n + m; i++) { double a; scanf("%lf", &a); addedge(i, t, log(a)); } for (int i = 0; i < num; i++) { int a, b; scanf("%d%d", &a, &b); addedge(a, n + b, inf); }}int main(){ //freopen("t.txt", "r", stdin); int case_num; scanf("%d", &case_num); while (case_num--) { ne = 2; memset(head, 0, sizeof(head)); input(); printf("%.4f\n", exp(flow(n + m + 2, s, t))); } return 0;}