POJ2112 Optimal Milking —— 二分圖多重匹配/最大流 + 二分
阿新 • • 發佈:2017-11-12
題目 none utili cnblogs tell pac log head star
Each milking point can "process" at most M (1 <= M <= 15) cows each day.
Write a program to find an assignment for each cow to some milking machine so that the distance the furthest-walking cow travels is minimized (and, of course, the milking machines are not overutilized). At least one legal assignment is possible for all input data sets. Cows can traverse several paths on the way to their milking machine.
* Line 1: A single line with three space-separated integers: K, C, and M.
* Lines 2.. ...: Each of these K+C lines of K+C space-separated integers describes the distances between pairs of various entities. The input forms a symmetric matrix. Line 2 tells the distances from milking machine 1 to each of the other entities; line 3 tells the distances from machine 2 to each of the other entities, and so on. Distances of entities directly connected by a path are positive integers no larger than 200. Entities not directly connected by a path have a distance of 0. The distance from an entity to itself (i.e., all numbers on the diagonal) is also given as 0. To keep the input lines of reasonable length, when K+C > 15, a row is broken into successive lines of 15 numbers and a potentially shorter line to finish up a row. Each new row begins on its own line.
A single line with a single integer that is the minimum possible total distance for the furthest walking cow.
題目鏈接:https://vjudge.net/problem/POJ-2112
Optimal Milking
Time Limit: 2000MS | Memory Limit: 30000K | |
Total Submissions: 18555 | Accepted: 6626 | |
Case Time Limit: 1000MS |
Description
FJ has moved his K (1 <= K <= 30) milking machines out into the cow pastures among the C (1 <= C <= 200) cows. A set of paths of various lengths runs among the cows and the milking machines. The milking machine locations are named by ID numbers 1..K; the cow locations are named by ID numbers K+1..K+C.Each milking point can "process" at most M (1 <= M <= 15) cows each day.
Write a program to find an assignment for each cow to some milking machine so that the distance the furthest-walking cow travels is minimized (and, of course, the milking machines are not overutilized). At least one legal assignment is possible for all input data sets. Cows can traverse several paths on the way to their milking machine.
Input
* Lines 2.. ...: Each of these K+C lines of K+C space-separated integers describes the distances between pairs of various entities. The input forms a symmetric matrix. Line 2 tells the distances from milking machine 1 to each of the other entities; line 3 tells the distances from machine 2 to each of the other entities, and so on. Distances of entities directly connected by a path are positive integers no larger than 200. Entities not directly connected by a path have a distance of 0. The distance from an entity to itself (i.e., all numbers on the diagonal) is also given as 0. To keep the input lines of reasonable length, when K+C > 15, a row is broken into successive lines of 15 numbers and a potentially shorter line to finish up a row. Each new row begins on its own line.
Output
Sample Input
2 3 2 0 3 2 1 1 3 0 3 2 0 2 3 0 1 0 1 2 1 0 2 1 0 0 2 0
Sample Output
2
Source
USACO 2003 U S Open
題解:
多重匹配:
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4View Code#include <cstdlib> 5 #include <string> 6 #include <vector> 7 #include <map> 8 #include <set> 9 #include <queue> 10 #include <sstream> 11 #include <algorithm> 12 using namespace std; 13 const int INF = 2e9; 14 const int MOD = 1e9+7; 15 const int MAXM = 5e2+10; 16 const int MAXN = 4e2+10; 17 18 int uN, vN, m, N, maze[MAXN][MAXN]; 19 int num[MAXM], linker[MAXM][MAXN]; 20 bool g[MAXN][MAXM], used[MAXM]; 21 22 bool dfs(int u) 23 { 24 for(int v = 0; v<vN; v++) 25 if(g[u][v] && !used[v]) 26 { 27 used[v] = true; 28 if(linker[v][0]<num[v]) 29 { 30 linker[v][++linker[v][0]] = u; 31 return true; 32 } 33 for(int i = 1; i<=num[v]; i++) 34 if(dfs(linker[v][i])) 35 { 36 linker[v][i] = u; 37 return true; 38 } 39 } 40 return false; 41 } 42 43 bool hungary(int mid) 44 { 45 memset(g, false, sizeof(g)); 46 for(int i = vN; i<N; i++) 47 for(int j = 0; j<vN; j++) 48 if(maze[i][j]<=mid) 49 g[i][j] = true; 50 51 for(int i = 0; i<vN; i++) 52 { 53 num[i] = m; 54 linker[i][0] = 0; 55 } 56 for(int u = vN; u<N; u++) 57 { 58 memset(used, false, sizeof(used)); 59 if(!dfs(u)) return false; 60 } 61 return true; 62 } 63 64 void Flyod() 65 { 66 for(int k = 0; k<N; k++) 67 for(int i = 0; i<N; i++) 68 for(int j = 0; j<N; j++) 69 maze[i][j] = min(maze[i][j], maze[i][k]+maze[k][j]); 70 } 71 72 int main() 73 { 74 while(scanf("%d%d%d", &vN, &uN, &m)!=EOF) 75 { 76 N = uN + vN; 77 for(int i = 0; i<N; i++) 78 for(int j = 0; j<N; j++) 79 { 80 scanf("%d", &maze[i][j]); 81 if(maze[i][j]==0) maze[i][j] = INF/2; 82 } 83 84 Flyod(); 85 int l = 1, r = 200*400; 86 while(l<=r) 87 { 88 int mid = (l+r)>>1; 89 if(hungary(mid)) 90 r = mid - 1; 91 else 92 l = mid + 1; 93 } 94 printf("%d\n", l); 95 } 96 }
最大流:
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <string> 6 #include <vector> 7 #include <map> 8 #include <set> 9 #include <queue> 10 #include <sstream> 11 #include <algorithm> 12 using namespace std; 13 const int INF = 2e9; 14 const int MOD = 1e9+7; 15 const int MAXM = 5e2+10; 16 const int MAXN = 4e2+10; 17 18 struct Edge 19 { 20 int to, next, cap, flow; 21 }edge[MAXN*MAXN]; 22 int tot, head[MAXN]; 23 24 int uN, vN, m, N, maze[MAXN][MAXN]; 25 int gap[MAXN], dep[MAXN], pre[MAXN], cur[MAXN]; 26 void add(int u, int v, int w) 27 { 28 edge[tot].to = v; edge[tot].cap = w; edge[tot].flow = 0; 29 edge[tot].next = head[u]; head[u] = tot++; 30 edge[tot].to = u; edge[tot].cap = 0; edge[tot].flow = 0; 31 edge[tot].next = head[v]; head[v] = tot++; 32 } 33 34 int sap(int start, int end, int nodenum) 35 { 36 memset(gap,0,sizeof(gap)); 37 memset(dep,0,sizeof(dep)); 38 memcpy(cur,head,sizeof(head)); 39 40 int u = start; 41 pre[u] = -1; 42 gap[0] = nodenum; 43 int maxflow = 0; 44 while(dep[start]<nodenum) 45 { 46 bool flag = false; 47 for(int i = cur[u]; i!=-1; i=edge[i].next) 48 { 49 int v = edge[i].to; 50 if(edge[i].cap-edge[i].flow && dep[v]+1==dep[u]) 51 { 52 flag = true; 53 cur[u] = pre[v] = i; 54 u = v; 55 break; 56 } 57 } 58 59 if(flag) 60 { 61 if(u==end) 62 { 63 int minn = INF; 64 for(int i = pre[u]; i!=-1; i=pre[edge[i^1].to]) 65 if(minn>edge[i].cap-edge[i].flow) 66 minn = edge[i].cap-edge[i].flow; 67 for(int i = pre[u]; i!=-1; i=pre[edge[i^1].to]) 68 { 69 edge[i].flow += minn; 70 edge[i^1].flow -= minn; 71 } 72 u = start; 73 maxflow += minn; 74 } 75 } 76 77 else 78 { 79 int minn = nodenum; 80 for(int i = head[u]; i!=-1; i=edge[i].next) 81 if(edge[i].cap-edge[i].flow && dep[edge[i].to]<minn) 82 { 83 minn = dep[edge[i].to]; 84 cur[u] = i; 85 } 86 gap[dep[u]]--; 87 if(gap[dep[u]]==0) break; 88 dep[u] = minn+1; 89 gap[dep[u]]++; 90 if(u!=start) u = edge[pre[u]^1].to; 91 } 92 } 93 return maxflow; 94 } 95 96 bool test(int mid) 97 { 98 tot = 0; 99 memset(head, -1, sizeof(head)); 100 for(int i = vN; i<N; i++) 101 { 102 add(N, i, 1); 103 for(int j = 0; j<vN; j++) 104 if(maze[i][j]<=mid) 105 add(i, j, 1); 106 } 107 for(int i = 0; i<vN; i++) 108 add(i, N+1, m); 109 110 int maxflow = sap(N, N+1, N+2); 111 return maxflow == uN; 112 } 113 114 void Flyod() 115 { 116 for(int k = 0; k<N; k++) 117 for(int i = 0; i<N; i++) 118 for(int j = 0; j<N; j++) 119 maze[i][j] = min(maze[i][j], maze[i][k]+maze[k][j]); 120 } 121 122 int main() 123 { 124 while(scanf("%d%d%d", &vN, &uN, &m)!=EOF) 125 { 126 N = uN + vN; 127 for(int i = 0; i<N; i++) 128 for(int j = 0; j<N; j++) 129 { 130 scanf("%d", &maze[i][j]); 131 if(maze[i][j]==0) maze[i][j] = INF/2; 132 } 133 134 Flyod(); 135 int l = 1, r = 200*400; 136 while(l<=r) 137 { 138 int mid = (l+r)>>1; 139 if(test(mid)) 140 r = mid - 1; 141 else 142 l = mid + 1; 143 } 144 printf("%d\n", l); 145 } 146 }View Code
POJ2112 Optimal Milking —— 二分圖多重匹配/最大流 + 二分