POJ 2112 Optimal Milking （二分 + floyd + 网络流）

POJ 2112 Optimal Milking 

链接：http://poj.org/problem?id=2112

题意：农场主John 将他的K(1≤K≤30)个挤奶器运到牧场，在那里有C(1≤C≤200)头奶牛，在奶牛和挤奶器之间有一组不同长度的路。K个挤奶器的位置用1～K的编号标明，奶牛的位置用K+1～K+C 的编号标明。每台挤奶器每天最多能为M(1≤M≤15)头奶牛挤奶。寻找一个方案，安排每头奶牛到某个挤奶器挤奶，并使得C 头奶牛需要走的所有路程中的最大路程最小。每个测试数据中至少有一个安排方案。每条奶牛到挤奶器有多条路。

思路：先用Floyd 算法求出能达到的任意两点之间的最短路径，然后二分最大距离的最小值，每次用二分的值求最大流。

在求最大流时构图：建立一个源点，每个点到挤奶器连一条流量为m的边。建立一个汇点，每头奶牛到汇点连一条流量为1的边。挤奶器与奶牛之间的距离小于等于mid则连边，流量为1。最后求最大流是否为c即可。

代码：

/*ID: [e-mail]:LANG: C++*/#include<map>#include<set>#include<queue>#include<stack>#include<cmath>#include<cstdio>#include<vector>#include<string>#include<fstream>#include<cstring>#include<ctype.h>#include<iostream>#include<algorithm>using namespace std;#define INF (1<<30)#define PI acos(-1.0)#define mem(a, b) memset(a, b, sizeof(a))#define rep(i, a, n) for (int i = a; i < n; i++)#define per(i, a, n) for (int i = n - 1; i >= a; i--)#define eps 1e-6#define debug puts("===============")#define pb push_back//#define mp make_pair#define all(x) (x).begin(),(x).end()#define fi first#define se second#define SZ(x) ((int)(x).size())#define POSIN(x,y) (0 <= (x) && (x) < n && 0 <= (y) && (y) < m)typedef long long ll;typedef unsigned long long ULL;const int maxn = 250;const int maxm = 100000;int k, c, m;int mp[maxn][maxn];struct node {    int v;    // vertex    int cap;    // capacity    int flow;   // current flow in this arc    int nxt;} e[maxm * 2];int g[maxn], cnt;int st, ed, n;void add(int u, int v, int c) {    e[++cnt].v = v;    e[cnt].cap = c;    e[cnt].flow = 0;    e[cnt].nxt = g[u];    g[u] = cnt;    e[++cnt].v = u;    e[cnt].cap = 0;    e[cnt].flow = 0;    e[cnt].nxt = g[v];    g[v] = cnt;}void init(int mid) {    mem(g, 0);    cnt = 1;    st = 0, ed = k + c + 1;    n = k + c;    for (int i = 1; i <= k; i++) add(st, i, m);    for (int i = k + 1; i <= n; i++) {        for (int j = 1; j <= k; j++) if (mp[i][j] <= mid) add(j, i, 1);        add(i, ed, 1);    }    n += 3;}int dist[maxn], numbs[maxn], q[maxn];void rev_bfs() {    int font = 0, rear = 1;    for (int i = 0; i <= n; i++) { //n为总点数        dist[i] = maxn;        numbs[i] = 0;    }    q[font] = ed;    dist[ed] = 0;    numbs[0] = 1;    while(font != rear) {        int u = q[font++];        for (int i = g[u]; i; i = e[i].nxt) {            if (e[i ^ 1].cap == 0 || dist[e[i].v] < maxn) continue;            dist[e[i].v] = dist[u] + 1;            ++numbs[dist[e[i].v]];            q[rear++] = e[i].v;        }    }}int maxflow() {    rev_bfs();    int u, totalflow = 0;    int curg[maxn], revpath[maxn];    for(int i = 0; i <= n; ++i) curg[i] = g[i];    u = st;    while(dist[st] < n) {        if(u == ed) {   // find an augmenting path            int augflow = INF;            for(int i = st; i != ed; i = e[curg[i]].v)                augflow = min(augflow, e[curg[i]].cap);            for(int i = st; i != ed; i = e[curg[i]].v) {                e[curg[i]].cap -= augflow;                e[curg[i] ^ 1].cap += augflow;                e[curg[i]].flow += augflow;                e[curg[i] ^ 1].flow -= augflow;            }            totalflow += augflow;            u = st;        }        int i;        for(i = curg[u]; i; i = e[i].nxt)            if(e[i].cap > 0 && dist[u] == dist[e[i].v] + 1) break;        if(i) {   // find an admissible arc, then Advance            curg[u] = i;            revpath[e[i].v] = i ^ 1;            u = e[i].v;        } else {    // no admissible arc, then relabel this vertex            if(0 == (--numbs[dist[u]])) break;    // GAP cut, Important!            curg[u] = g[u];            int mindist = n;            for(int j = g[u]; j; j = e[j].nxt)                if(e[j].cap > 0) mindist = min(mindist, dist[e[j].v]);            dist[u] = mindist + 1;            ++numbs[dist[u]];            if(u != st)                u = e[revpath[u]].v;    // Backtrack        }    }    return totalflow;}void get() {    n = k + c;    for (int i = 1; i <= n; i++) {        for (int j = 1; j <= n; j++) {            scanf("%d", mp[i] + j);            if (mp[i][j] == 0 && i != j) mp[i][j] = INF;        }    }}void floyd(int n, int mp[][maxn]) {    for (int k = 1; k <= n; k++) {        for (int i = 1; i <= n; i++) if (mp[i][k] != INF) {            for (int j = 1; j <= n; j++) mp[i][j] = min(mp[i][j], mp[i][k] + mp[k][j]);        }    }}int main () {    while(scanf("%d%d%d", &k, &c, &m) != EOF) {        get();        floyd(n, mp);        int l = 0, r = 10000, mid;        while(l < r) {            mid = (l + r) >> 1;            init(mid);            //debug;            if (maxflow() >= c) r = mid;            else l = mid + 1;        }        printf("%d\n", r);    }    return 0;}