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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: wuqi9395@126.com PROG: 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; }
POJ 2112 Optimal Milking (二分 + floyd + 网络流)
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