POJ2112 Optimal Milking 【最大流+二分】
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POJ2112 Optimal Milking 【最大流+二分】
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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.
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.
* 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
Source
題意：有k臺擠奶器，每臺擠奶器最多容納m頭奶牛，該牧場共同擁有c頭奶牛，如今給定這k臺機器和c頭奶牛相互間的直接距離，求讓全部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，然后設置源點到每頭牛的距離為1，每臺機器到匯點的距離為m，然后若牛到機器的距離不大于maxdist，那么則將該邊增加到新圖中，最后對新圖求最大流，推斷最大流是否等于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 235 int 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;
}
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