floyd + 最大流 （奶牛分配问题）

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

题意 ： 有 k 台机器, c 头奶牛， 每台机器最多可供喂养的奶牛的数量 m , 开始给出任意两点之间的距离，你可以随意的安置奶牛，要求奶牛走的最远的距离最小
思路分析 ： 
　　首先可以用 floyd 跑出任意两点的最短路
　　然后就是一个网络流即可，用源点和每个机器建立一条边，流量为 m , 用汇点和每头牛建立一条边，流量为 1，并且二分答案，当机器和牛的距离小于 mid 时，即可连边，每次更新答案即可
代码示例 ：

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;
}