Paratroopers
It is year 2500 A.D. and there is a terrible war between the forces of the Earth and the Mars. Recently, the commanders of the Earth are informed by their spies that the invaders of Mars want to land some paratroopers in the m × n grid yard of one their main weapon factories in order to destroy it. In addition, the spies informed them the row and column of the places in the yard in which each paratrooper will land. Since the paratroopers are very strong and well-organized, even one of them, if survived, can complete the mission and destroy the whole factory. As a result, the defense force of the Earth must kill all of them simultaneously after their landing.
In order to accomplish this task, the defense force wants to utilize some of their most hi-tech laser guns. They can install a gun on a row (resp. column) and by firing this gun all paratroopers landed in this row (resp. column) will die. The cost of installing a gun in the ith row (resp. column) of the grid yard is ri (resp. ci ) and the total cost of constructing a system firing all guns simultaneously is equal to the product of their costs. Now, your team as a high rank defense group must select the guns that can kill all paratroopers and yield minimum total cost of constructing the firing system.
Input
Input begins with a number T showing the number of test cases and then, T test cases follow. Each test case begins with a line containing three integers 1 ≤ m ≤ 50 , 1 ≤ n ≤ 50 and 1 ≤ l ≤ 500 showing the number of rows and columns of the yard and the number of paratroopers respectively. After that, a line with m positive real numbers greater or equal to 1.0 comes where the ith number is ri and then, a line with n positive real numbers greater or equal to 1.0 comes where the ith number is ci. Finally, l lines come each containing the row and column of a paratrooper.
Output
For each test case, your program must output the minimum total cost of constructing the firing system rounded to four digits after the fraction point.
Sample Input
1
4 4 5
2.0 7.0 5.0 2.0
1.5 2.0 2.0 8.0
1 1
2 2
3 3
4 4
1 4
Sample Output
16.0000
Source
Amirkabir University of Technology Local Contest 2006
题目类型：最小割
算法分析：这是典型的顶点覆盖问题。常见的建模方式是将横行和列分别看作是顶点，从源点向每个横行点连一条容量为该行消灭敌人的cost，每个列点向汇点连一条容量为该列消灭敌人的cost。最后对于每个火星人的位置，从对应的行向对应的列连一条容量为INF的有向边，最后跑个最大流即可。注意这里为处理问题方便，将每个点容量取对数后计算，%lf在G++下会WA！
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/************************************************* Author :supermaker Created Time :2016/3/30 18:15:39 File Location :C:\Users\abcd\Desktop\TheEternalPoet **************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7F7F7F7F; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 200 + 66; struct Node { int to, nxt; double c, f; }; Node edge[maxn*maxn]; int head[maxn], edgelen; int dep[maxn], cnt[maxn], cur[maxn]; void Init () { memset (head, -1, sizeof (head)); memset (edge, -1, sizeof (edge)); edgelen = 0; } void AddEdge (int u, int v, double w, double rw = 0) { edge[edgelen].to = v, edge[edgelen].c = w, edge[edgelen].f = 0; edge[edgelen].nxt = head[u], head[u] = edgelen++; edge[edgelen].to = u, edge[edgelen].c = rw, edge[edgelen].f = 0; edge[edgelen].nxt = head[v], head[v] = edgelen++; } void bfs (int s, int e) { memset (dep, -1, sizeof (dep)); memset (cnt, 0, sizeof (cnt)); dep[e] = 0; cnt[dep[e]]++; queue <int> qu; qu.push (e); while (!qu.empty ()) { int u = qu.front (); qu.pop (); for (int i = head[u]; i != -1; i = edge[i].nxt) { int v = edge[i].to; if (dep[v] != -1) continue; qu.push (v); dep[v] = dep[u] + 1; cnt[dep[v]]++; } } } int sta[maxn]; double MaxFlow (int s, int e, int n) { bfs (s, e); double res = 0; int top = 0, u = s; memcpy (cur, head, sizeof (head)); while (dep[s] < n) { if (u == e) { double minval = INF; int minval_pos = -1; for (int i = 0; i < top; i++) if (minval > edge[sta[i]].c - edge[sta[i]].f) { minval = edge[sta[i]].c - edge[sta[i]].f; minval_pos = i; } for (int i = 0; i < top; i++) { edge[sta[i]].f += minval; edge[sta[i]^1].f -= minval; } res += minval; top = minval_pos; u = edge[sta[top]^1].to; } else { bool is_have = false; for (int i = cur[u]; i != -1; i = edge[i].nxt) { int v = edge[i].to; if (fabs (edge[i].c - edge[i].f) > EPS && dep[v] + 1 == dep[u]) { cur[u] = i; sta[top++] = cur[u]; u = v; is_have = true; break; } } if (is_have) continue; int minval = n; for (int i = head[u]; i != -1; i = edge[i].nxt) if (fabs (edge[i].c - edge[i].f) > EPS && dep[edge[i].to] < minval) { minval = dep[edge[i].to]; cur[u] = i; } cnt[dep[u]]--; if (!cnt[dep[u]]) return res; dep[u] = minval + 1; cnt[dep[u]]++; if (u != s) u = edge[sta[--top]^1].to; } } return res; } int main() { //CFF; //CPPFF; int t; scanf ("%d", &t); while (t--) { Init (); int row, col, num; scanf ("%d%d%d", &row, &col, &num); for (int i = 1; i <= row; i++) { double v; scanf ("%lf", &v); AddEdge (0, i, log (v)); } for (int i = 1; i <= col; i++) { double v; scanf ("%lf", &v); AddEdge (i + row, row + col + 1, log (v)); } for (int i = 1; i <= num; i++) { int x, y; scanf ("%d%d", &x, &y); AddEdge (x, row + y, INF); } printf ("%.4lf\n", exp (MaxFlow (0, row + col + 1, row + col + 2))); } return 0; } |