描述
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.
输入
* 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.
输出
A single line with a single integer that is the minimum possible total distance for the furthest walking cow.
样例输入
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
样例输出
2
题意
给你(K+C)*(K+C)的图，K个牛奶机，每个牛奶机最多供M头牛，一共C头牛，问所有方案中使得距离牛奶机器最远的牛的距离最小
题解
先把牛奶机连汇点T流量M，牛连源点S流量1，牛和牛奶机连流量1，如果C牛都能有饮料机，则说明汇点T=C
然后是怎么连牛和牛奶机的问题，可以知道答案求的是最大值最小
直接二分答案[0,200*(K+C)]
每次把距离<=mid的边加进去，如果T=C，则说明可行，r=mid
否则l=mid
代码
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 const int maxn=1e5+5; 5 const int maxm=2e5+5; 6 int n,m,S,T; 7 int deep[maxn],q[400000]; 8 int FIR[maxn],TO[maxm],CAP[maxm],COST[maxm],NEXT[maxm],tote; 9 10 void add(int u,int v,int cap) 11 { 12 TO[tote]=v; 13 CAP[tote]=cap; 14 NEXT[tote]=FIR[u]; 15 FIR[u]=tote++; 16 17 TO[tote]=u; 18 CAP[tote]=0; 19 NEXT[tote]=FIR[v]; 20 FIR[v]=tote++; 21 } 22 bool bfs() 23 { 24 memset(deep,0,sizeof deep); 25 deep[S]=1;q[1]=S; 26 int head=0,tail=1; 27 while(head!=tail) 28 { 29 int u=q[++head]; 30 for(int v=FIR[u];v!=-1;v=NEXT[v]) 31 { 32 if(CAP[v]&&!deep[TO[v]]) 33 { 34 deep[TO[v]]=deep[u]+1; 35 q[++tail]=TO[v]; 36 } 37 } 38 } 39 return deep[T]; 40 } 41 int dfs(int u,int fl) 42 { 43 if(u==T)return fl; 44 int f=0; 45 for(int v=FIR[u];v!=-1&&fl;v=NEXT[v]) 46 { 47 if(CAP[v]&&deep[TO[v]]==deep[u]+1) 48 { 49 int Min=dfs(TO[v],min(fl,CAP[v])); 50 CAP[v]-=Min;CAP[v^1]+=Min; 51 fl-=Min;f+=Min; 52 } 53 } 54 if(!f)deep[u]=-2; 55 return f; 56 } 57 int maxflow() 58 { 59 int ans=0; 60 while(bfs()) 61 ans+=dfs(S,1<<30); 62 return ans; 63 } 64 void init() 65 { 66 tote=0; 67 memset(FIR,-1,sizeof FIR); 68 } 69 int K,C,N,M,a[250][250]; 70 int main() 71 { 72 cin>>K>>C>>M; 73 N=K+C; 74 for(int i=1;i<=N;i++) 75 for(int j=1;j<=N;j++) 76 { 77 scanf("%d",&a[i][j]); 78 if(i!=j&&!a[i][j])a[i][j]=0x3f3f3f3f; 79 } 80 for(int k=1;k<=N;k++) 81 for(int i=1;i<=N;i++) 82 for(int j=1;j<=N;j++) 83 if(a[i][j]>a[i][k]+a[k][j]) 84 a[i][j]=a[i][k]+a[k][j]; 85 int l=0,r=200*N; 86 S=0,T=K+C+1; 87 while(r-l>1) 88 { 89 int mid=(l+r)>>1; 90 init(); 91 for(int i=1;i<=K;i++) 92 add(S,i,M); 93 for(int i=K+1;i<=N;i++) 94 add(i,T,1); 95 for(int i=1;i<=K;i++) 96 for(int j=K+1;j<=N;j++) 97 if(a[i][j]&&a[i][j]<=mid) 98 add(i,j,a[i][j]); 99 int sum=maxflow(); 100 if(sum==C)r=mid; 101 else l=mid; 102 } 103 printf("%d ",r); 104 return 0; 105 }