POJ 3308 Paratroopers
题意：有一个N*M的方阵，有L个伞兵降落在方阵上。现在要将所有的伞兵都消灭掉，可以在每行每列装一个高射炮，如果在某行（某列）装上高射炮之后，能够消灭所有落在该行（该列）的伞兵。每行每列安高射炮有费用，问如何安装能够使得费用之积最小。
思路：首先题目要求乘积最小，将乘积对e取对数，会发现就变成了求和。然后抽象出一个二分图，每一行是x部的一个点，每个点有权值，权值为费用取ln。每一列是y部的一点，费用计算相同。如果有伞兵降落在方格上，那么将x部与y部连边。问题就成了求该二分图的最小点权覆盖。通过求最小割即可得到。而最小割就是最大流。所以构造一个流图。建立一个源点，源点向每一行连边，流量为费用，建立一个汇点，每列向汇点连边，流量为费用。二分图中原先的边的流量是INF。最后答案就是exp（最大流）。
细节：INF不能取太大，否则精度会有问题。
代码：
/*
ID: wuqi9395@126.com
PROG:
LANG: C++
*/
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define LINF (1LL<<60)
#define INF 1e8
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i < n; i++)
#define per(i, a, n) for (int i = n - 1; i >= a; i--)
#define eps 1e-6
#define debug puts("===============")
#define pb push_back
//#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define POSIN(x,y) (0 <= (x) && (x) < n && 0 <= (y) && (y) < m)
typedef long long ll;
typedef unsigned long long ULL;
const int maxn = 110;
const int maxm = 20000;
int st, ed, n, m, l;
struct node {
int v; // vertex
double cap; // capacity
double flow; // current flow in this arc
int nxt;
} e[maxm * 2];
int g[maxn], cnt;
void add(int u, int v, double c) {
e[++cnt].v = v;
e[cnt].cap = c;
e[cnt].flow = 0;
e[cnt].nxt = g[u];
g[u] = cnt;
e[++cnt].v = u;
e[cnt].cap = 0;
e[cnt].flow = 0;
e[cnt].nxt = g[v];
g[v] = cnt;
}
void init() {
mem(g, 0);
cnt = 1;
st = 0;
ed = m + n + 1;
double w;
int u, v;
for (int i = 1; i <= m; i++) {
scanf("%lf", &w);
add(st, i, log(w));
}
for (int i = m + 1; i < ed; i++) {
scanf("%lf", &w);
add(i, ed, log(w));
}
for (int i = 1; i <= l; i++) {
scanf("%d%d", &u, &v);
add(u, m + v, INF);
}
n = ed + 3;
}
int dist[maxn], numbs[maxn], q[maxn];
void rev_bfs() {
int font = 0, rear = 1;
for (int i = 0; i <= n; i++) { //n为总点数
dist[i] = maxn;
numbs[i] = 0;
}
q[font] = ed;
dist[ed] = 0;
numbs[0] = 1;
while(font != rear) {
int u = q[font++];
for (int i = g[u]; i; i = e[i].nxt) {
if (e[i ^ 1].cap == 0 || dist[e[i].v] < maxn) continue;
dist[e[i].v] = dist[u] + 1;
++numbs[dist[e[i].v]];
q[rear++] = e[i].v;
}
}
}
double maxflow() {
rev_bfs();
int u;
double totalflow = 0;
int curg[maxn], revpath[maxn];
for(int i = 0; i <= n; ++i) curg[i] = g[i];
u = st;
while(dist[st] < n) {
if(u == ed) { // find an augmenting path
double augflow = INF;
for(int i = st; i != ed; i = e[curg[i]].v)
augflow = min(augflow, e[curg[i]].cap);
for(int i = st; i != ed; i = e[curg[i]].v) {
e[curg[i]].cap -= augflow;
e[curg[i] ^ 1].cap += augflow;
e[curg[i]].flow += augflow;
e[curg[i] ^ 1].flow -= augflow;
}
totalflow += augflow;
u = st;
}
int i;
for(i = curg[u]; i; i = e[i].nxt)
if(e[i].cap > 0 && dist[u] == dist[e[i].v] + 1) break;
if(i) { // find an admissible arc, then Advance
curg[u] = i;
revpath[e[i].v] = i ^ 1;
u = e[i].v;
} else { // no admissible arc, then relabel this vertex
if(0 == (--numbs[dist[u]])) break; // GAP cut, Important!
curg[u] = g[u];
int mindist = n;
for(int j = g[u]; j; j = e[j].nxt)
if(e[j].cap > 0) mindist = min(mindist, dist[e[j].v]);
dist[u] = mindist + 1;
++numbs[dist[u]];
if(u != st)
u = e[revpath[u]].v; // Backtrack
}
}
return totalflow;
}
int main () {
int T;
scanf("%d", &T);
while(T--) {
scanf("%d%d%d", &m, &n, &l);
init();
printf("%.4f\n", exp(maxflow()));
}
return 0;
}