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 * 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.
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

題意 ： 有 k 臺機器, c 頭奶牛， 每臺機器最多可供餵養的奶牛的數量 m , 開始給出任意兩點之間的距離，你可以隨意的安置奶牛，要求奶牛走的最遠的距離最小
思路分析 ： 
　　首先可以用 floyd 跑出任意兩點的最短路
　　然後就是一個網路流即可，用源點和每個機器建立一條邊，流量為 m , 用匯點和每頭牛建立一條邊，流量為 1，並且二分答案，當機器和牛的距離小於 mid 時，即可連邊，每次更新答案即可
程式碼示例 ：

const int maxn = 1e4+5;
const int inf = 0x3f3f3f3f;

int k, c, m;
struct node
{
    int to, next;
    int flow;
}e[maxn<<1];
int head[maxn];
int cnt;
int mp[300][300];

void addedge(int u, int v, int w){
    e[cnt].to = v, e[cnt].flow = w, e[cnt].next = head[u], head[u] = cnt++;
    e[cnt].to = u, e[cnt].flow = 0, e[cnt].next = head[v], head[v] = cnt++;
}

void flod() {
    int sum = k+c;
    for(int f = 1; f <= sum; f++){
        for(int i = 1; i <= sum; i++){
            for(int j = 1; j <= sum; j++){
                mp[i][j] = min(mp[i][f]+mp[f][j], mp[i][j]); 
            }
        }
    }
}   

int dep[maxn], que[maxn];
bool bfs(int s, int t){
    memset(dep, 0, sizeof(dep));
    int head1 = 0, tail = 1; dep[s] = 1;
    que[0] = s;
    while(head1 < tail) {
        int u = que[head1++]; 
        for(int i = head[u]; i != -1; i = e[i].next) {
            int to = e[i].to;
            if (e[i].flow && !dep[to]) {
                dep[to] = dep[u]+1;
                que[tail++] = to;
            }
        }
    }
    return dep[t];
}
int aim;
int dfs(int u, int f1){
    if (u == aim || f1 == 0) return f1;
    
    int f = 0;
    for(int i = head[u]; i != -1; i = e[i].next){
        int to = e[i].to;
        if (e[i].flow && dep[to] == dep[u]+1){
            int x = dfs(to, min(f1, e[i].flow));
            e[i].flow -= x; e[i^1].flow += x; 
            f1 -= x, f += x;
            if (f1 == 0) return f;
        }
    }
    if (!f) dep[u] = -2;
    return f;
}

bool check(int lenth){
    int s = 0, t = k+c+1;
    memset(head, -1, sizeof(head));
    cnt = 0; aim = t;
    
    for(int i = 1; i <= k; i++) addedge(s, i, m);
    for(int i = k+1; i <= k+c; i++) addedge(i, t, 1);
    for(int i = 1; i <= k; i++){
        for(int j = k+1; j <= k+c; j++){
            if (mp[i][j] <= lenth) addedge(i, j, 1);
        }
    }
    int res = 0;
    while(bfs(s, t)){
        res += dfs(s, inf); 
    }
    if (res == c) return true;
    return false;
}

int main() {
    int x;
    
    cin >> k >> c >> m;
    memset(mp, inf, sizeof(mp));
    for(int i = 1; i <= k+c; i++){
        for(int j = 1; j <= k+c; j++){
            scanf("%d", &x); 
            if (x != 0) mp[i][j] = x;    
        }
    }
    flod(); 
    int l = 0, r = 1e5;
    int ans;
    while(l <= r){
        int mid = (l+r)>>1;
        if (check(mid)) {ans = mid; r = mid-1;}
        else l = mid+1;
    }
    cout <<ans << endl;
    return 0;
}