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 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.
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.
1 #include<cmath> 2 #include<iostream> 3 #include<algorithm> 4 #include<string> 5 #include<cstring> 6 #include<cstdio> 7 #define MAX 1000 8 #define eps 0.0001 9 #define INF 9999999999 10 #define min(a,b) (a<b?a:b) 11 using namespace std; 12 struct Edge 13 { 14 int st, ed; 15 int next; 16 double flow; 17 } edge[MAX*10]; 18 19 int pre[MAX]; // pre[u]储存以u为终点的边的编号，它储存的是边 20 int head[MAX], out[MAX]; 21 int leve[MAX]; // 层次图中每个点的层次 22 int que[MAX], stk[MAX]; 23 int E, R, C, L, src, dest; 24 25 void add_edge ( int u, int v, double val ) 26 { 27 edge[E].st = u; 28 edge[E].ed = v; 29 edge[E].flow = val; 30 edge[E].next = head[u]; 31 head[u] = E++; 32 33 edge[E].st = v; 34 edge[E].ed = u; 35 edge[E].flow = 0; 36 edge[E].next = head[v]; 37 head[v] = E++; 38 } 39 40 bool dinic_bfs () 41 { 42 int front, rear, u, v, i; 43 front = rear = 0; 44 memset(leve,-1,sizeof(leve)); 45 que[rear++] = src; 46 leve[src] = 0; 47 while ( front != rear ) 48 { 49 u = que[front]; 50 front = ( front + 1 ) % MAX; 51 for ( i = head[u]; i != -1; i = edge[i].next ) 52 { 53 v = edge[i].ed; 54 if ( leve[v] == -1 && edge[i].flow > eps ) 55 { 56 leve[v] = leve[u] + 1; 57 que[rear] = v; 58 rear = ( rear + 1 ) % MAX; 59 } 60 } 61 } 62 return leve[dest] >= 0; // leve[dest] == -1 说明没有找到增广路径 63 64 } 65 66 double dinic_dfs () 67 { 68 int top = 0, u, v, pos,i; 69 double minf, res = 0; 70 memcpy(out,head,sizeof(head)); // 将head的值拷给out 71 stk[top++] = u = src; 72 while ( top != 0 ) //栈空说明在当前层次图中，所有的增广路径寻找完毕 73 { 74 while ( u != dest ) 75 { 76 for ( i = out[u]; i != -1; i = edge[i].next ) 77 { 78 v = edge[i].ed; 79 if ( edge[i].flow > eps && leve[u] + 1 == leve[v] ) 80 { 81 out[u] = edge[i].next; // 下一次访问的时候，就可以直接得到edge[i].next 82 pre[v] = i; // 存储边的编号 83 stk[top++] = v; // 栈中放的是点 84 u = v; 85 break; 86 } 87 } 88 if ( i == -1 ) // 若从某点出发找不到合法的边 89 { 90 top--; // 把该点从栈中删除 91 if ( top <= 0 ) break; 92 u = stk[top-1]; // 返回前一个点 93 } 94 } 95 96 if ( u == dest ) 97 { 98 minf = INF; 99 while ( u != src ) 100 { 101 minf = min ( edge[pre[u]].flow, minf ); 102 u = edge[pre[u]].st; 103 } 104 105 u = dest; 106 while ( u != src ) 107 { 108 edge[pre[u]].flow -= minf; 109 edge[pre[u]^1].flow += minf; 110 if ( edge[pre[u]].flow < eps ) pos = edge[pre[u]].st; // 记录下零流出现的顶点。取最前面的哪一个。 111 u = edge[pre[u]].st; 112 } 113 114 while ( top > 0 && stk[top-1] != pos ) top--; // 退回到零流出现的顶点 115 if ( top > 0 ) u = stk[top-1]; 116 res += minf; 117 } 118 } 119 return res; 120 } 121 122 123 double Dinic () 124 { 125 double ans = 0, temp; 126 while ( dinic_bfs() ) 127 { 128 temp = dinic_dfs(); 129 if ( temp > eps ) ans+=temp; 130 else break; 131 } 132 return ans; 133 } 134 135 int main() 136 { 137 int T, a, b, i; 138 double val; 139 scanf("%d",&T); 140 while ( T-- ) 141 { 142 scanf("%d%d%d",&R,&C,&L); 143 memset(head,-1,sizeof(head)); 144 src = E = 0; 145 dest = R + C + 1; 146 for ( i = 1; i <= R; i++ ) 147 { 148 scanf("%lf",&val); 149 add_edge ( src, i, log(val) ); 150 } 151 152 for ( i = 1; i <= C; i++ ) 153 { 154 scanf("%lf",&val); 155 add_edge ( i+R, dest, log(val) ); 156 } 157 158 while ( L-- ) 159 { 160 scanf("%d%d",&a,&b); 161 add_edge(a,b+R,INF); 162 } 163 val = Dinic(); 164 printf("%.4f ",exp(val)); 165 } 166 return 0; 167 }