POJ2112 Optimal Milking 【最大流+二分】，poj2112milking

POJ2112 Optimal Milking 【最大流+二分】，poj2112milking

Optimal Milking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 12482		Accepted: 4508
Case Time Limit: 1000MS

Description

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 20 3 2 1 13 0 3 2 02 3 0 1 01 2 1 0 21 0 0 2 0

Sample Output

2

Source

USACO 2003 U S Open題意：有k台擠奶器，每台擠奶器最多容納m頭奶牛，該牧場共有c頭奶牛，現在給定這k台機器和c頭奶牛相互間的直接距離，求讓所有奶牛到達擠奶器且滿足該條件時奶牛走到擠奶器間的最大距離的最小值。題解：構圖：先用Floyd求出相互間的最短距離，然後設定源點到每頭牛的距離為1，每台機器到匯點的距離為m，然後若牛到機器的距離不大於maxdist，那麼則將該邊加入到新圖中，最後對新圖求最大流，判斷最大流是否等於c，就這樣二分枚舉maxdist直到找到最小的maxdist為止。

#include <stdio.h>#include <string.h>#define inf 0x3fffffff#define maxn 235int dist[maxn][maxn], k, c, m, n;int G[maxn][maxn], Layer[maxn];int queue[maxn], maxDist;bool vis[maxn];void Floyd() {    int x, i, j;    maxDist = 200;    for(x = 1; x <= n; ++x)        for(i = 1; i <= n; ++i)            for(j = 1; j <= n; ++j)                if(dist[i][j] > dist[i][x] + dist[x][j]) {                    dist[i][j] = dist[i][x] + dist[x][j];                    if(maxDist < dist[i][j]) maxDist = dist[i][j];                }}void build(int flow) {    memset(G, 0, sizeof(G));    int i, j;    for(i = k + 1; i <= n; ++i) {        G[0][i] = 1;        for(j = 1; j <= k; ++j)            if(dist[i][j] <= flow)                G[i][j] = 1;    }    for(j = 1; j <= k; ++j)        G[j][n + 1] = m;}bool countLayer() {    int id = 0, front = 0, now, i;    memset(Layer, 0, sizeof(Layer));    Layer[0] = 1; queue[id++] = 0;    while(front < id) {        now = queue[front++];        for(i = 0; i <= n + 1; ++i)            if(G[now][i] && !Layer[i]) {                Layer[i] = Layer[now] + 1;                if(i == n + 1) return true;                else queue[id++] = i;            }    }    return false;}bool Dinic() {    int i, maxFlow = 0, id = 0, now, minCut, pos, u, v;    while(countLayer()) {        memset(vis, 0, sizeof(vis));        vis[0] = 1; queue[id++] = 0;        while(id) {            now = queue[id - 1];            if(now == n + 1) {                minCut = inf;                for(i = 1; i < id; ++i) {                    u = queue[i - 1];                    v = queue[i];                    if(G[u][v] < minCut) {                        minCut = G[u][v];                        pos = u;                    }                 }                maxFlow += minCut;                for(i = 1; i < id; ++i) {                    u = queue[i - 1];                    v = queue[i];                    G[u][v] -= minCut;                    G[v][u] += minCut;                }                while(id && queue[id - 1] != pos)                    vis[queue[--id]] = 0;            } else {                for(i = 0; i <= n + 1; ++i) {                    if(G[now][i] && !vis[i] && Layer[now] + 1 == Layer[i]) {                        queue[id++] = i;                        vis[i] = 1; break;                    }                }                if(i > n + 1) --id;            }        }    }    return maxFlow == c;}int binarySolve() {    int left = 0, right = maxDist, mid;    while(left < right) {        mid = (left + right) >> 1;        build(mid);        if(Dinic()) right = mid;        else left = mid + 1;    }    return left;}int main() {    //freopen("stdin.txt", "r", stdin);    int i, j;    while(scanf("%d%d%d", &k, &c, &m) == 3) {        for(i = 1, n = k + c; i <= n; ++i)            for(j = 1; j <= n; ++j) {                scanf("%d", &dist[i][j]);                if(!dist[i][j] && i != j)                    dist[i][j] = inf;            }        Floyd();        printf("%d\n", binarySolve());    }    return 0;}