Optimal Milking
| Time Limit: 2000MS | Memory Limit: 30000K | |
| Total Submissions: 16550 | Accepted: 5945 | |
| 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.
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
* 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.
* 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
A single line with a single integer that is the minimum possible total distance for the furthest walking cow.
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
水平太差，做这个题可谓是历经千辛万苦，差不多做了两天多。开始想要偷懒，一次见图，再DINIC种根据边的大小进行判断是否可以搜索，失败！然后改为每次建图，然后dinic,然后就OK了，提交的时候有出了个低级失误，在POJ提交时用的C++，结果提示“
Main.cpp
xlocale(1242) : fatal error C1088: Cannot flush compiler intermediate file: '_CL_75c8ace5ex': No space left on device
”磁盘空间不足，但是就懵了，尝试了N次才发现，我的智商啊！！！！题目：floyd算出牛和挤奶器相互之间的最短距离，二分答案用最大流进行判断，如果最大流==牛数就可能是答案，否则不是。
代码如下：
1 Source Code 2 #include<cstdio> 3 #include<iostream> 4 #include<cstring> 5 #include<vector> 6 #include<queue> 7 8 using namespace std; 9 int k,c,m,ans; 10 int map[240][240]; 11 int tu[240][240]; 12 int lays[240]; 13 int vis[240]; 14 void floyd() 15 { 16 for(int kk=1;kk<=k+c;kk++) 17 for(int i=1;i<=k+c;i++) 18 for(int j=1;j<=k+c;j++) 19 if(map[i][kk]+map[kk][j]<map[i][j]) 20 map[i][j]=map[i][kk]+map[kk][j]; 21 } 22 void mideg(int mid) 23 { 24 memset(tu,0,sizeof(tu)); 25 for(int i=1;i<=k;i++) 26 { 27 tu[0][i]=m; 28 map[0][i]=1; 29 } 30 for(int i=1;i<=k;i++) 31 { 32 for(int j=k+1;j<=k+c;j++) 33 if(map[i][j]<=mid) 34 tu[i][j]=1; 35 } 36 for(int i=k+1;i<=k+c;i++) 37 { 38 tu[i][k+c+1]=1; 39 map[i][k+c+1]=1; 40 } 41 } 42 bool bfs() 43 { 44 memset(lays,-1,sizeof(lays)); 45 queue<int>q; 46 q.push(0); 47 lays[0]=0; 48 while(!q.empty()) 49 { 50 int u=q.front(); 51 q.pop(); 52 for(int i=0;i<=k+c+1;i++) 53 { 54 if(tu[u][i]>0&&lays[i]==-1) 55 { 56 lays[i]=lays[u]+1; 57 if(i==k+c+1)return 1; 58 else q.push(i); 59 } 60 } 61 } 62 return 0; 63 } 64 bool dinic() 65 { 66 int maxf=0; 67 vector<int>q; 68 while(bfs()) 69 { 70 memset(vis,0,sizeof(vis)); 71 q.push_back(0); 72 vis[0]=1; 73 while(!q.empty()) 74 { 75 int nd=q.back(); 76 if(nd==k+c+1) 77 { 78 int minn,minx=0x7fffffff; 79 for(int i=1;i<q.size();i++) 80 { 81 int u=q[i-1],v=q[i]; 82 if(minx>tu[u][v]) 83 { 84 minx=tu[u][v]; 85 minn=u; 86 } 87 } 88 maxf+=minx; 89 for(int i=1;i<q.size();i++) 90 { 91 int u=q[i-1],v=q[i]; 92 tu[u][v]-=minx; 93 tu[v][u]+=minx; 94 } 95 while(!q.empty()&&q.back()!=minn) 96 { 97 vis[q.back()]=0; 98 q.pop_back(); 99 } 100 } 101 else 102 { 103 int i; 104 for(i=0;i<=k+c+1;i++) 105 { 106 if(tu[nd][i]>0&&!vis[i]&&lays[i]==lays[nd]+1) 107 { 108 q.push_back(i); 109 vis[i]=1; 110 break; 111 } 112 } 113 if(i>k+c+1)q.pop_back(); 114 } 115 } 116 } 117 return maxf==c; 118 } 119 int main() 120 { 121 122 cin>>k>>c>>m; 123 memset(map,0x3f,sizeof(map)); 124 for(int i=1;i<=k+c;i++) 125 for(int j=1;j<=k+c;j++) 126 { 127 int a; 128 scanf("%d",&a); 129 if(a)map[i][j]=a; 130 if(i==j)map[i][j]=0; 131 } 132 133 floyd(); 134 135 int l=0,r=50000; 136 while(l<=r) 137 { 138 int mid=(l+r)/2; 139 mideg(mid); 140 bool pd=dinic(); 141 if(pd) 142 { 143 ans=mid; 144 r=mid-1; 145 } 146 else 147 l=mid+1; 148 } 149 cout<<ans<<endl; 150 return 0; 151 }