poj 2112 最大流+二分法

链接

http://poj.org/problem?id=2112

题意

安排C头牛到某个挤奶器，使得每头牛需要走的路程中最大路程的距离最小。

解析

先用floyd算法，求出奶牛到每个挤奶器的最短距离，然后从最大的距离二分，当此距离可以使C头牛，匹配成功，就让右区间r=mid;若不成功，就让左区间l=mid;之后继续二分。每次二分后，都要重新建图，另外加超级源点s和超级汇点t，让每头牛与汇点建边，容量为1，每个挤奶器与源点建边，容量为M，挤奶器与奶牛的距离<=mid，就建容量为1的边。之后用最大流算法，即可。

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include<queue>
#include <algorithm>
#include <map>
#include <set>
using namespace std;
const int maxn = 500+10;
const int inf = 0x3f3f3f3f;
int dis[maxn][maxn];
int level[maxn];
int n, K, C, M;
struct node {
int t;
int v;
int next;
}edge[maxn*maxn];
int head[maxn];
int cnt=0;
void init() {
cnt = 0;
memset(head, -1, sizeof(head));
}

void add(int u, int v, int c) {
edge[cnt].t = v;
edge[cnt].v = c;
edge[cnt].next = head[u];
head[u] = cnt;
cnt++;
edge[cnt].t = u;
edge[cnt].v = 0;
edge[cnt].next = head[v];
head[v] = cnt;
cnt++;
}

void build(int Dis) {
int s = 0;
int t = n+1;
init();
for (int i=K+1; i<=n; i++)
add(i, t, 1);
for (int i=1; i<=K; i++)
add(s, i, M);
for (int i=1; i<=K; i++) {
for (int j=K+1; j<=n; j++)
if (dis[i][j] <= Dis)
add(i, j, 1);
}
}

bool bfs() {
int q[maxn], front=0, tail = 0;
memset(level, -1, sizeof(level));
level[0] = 1;
q[tail++] = 0;
while (tail > front) {
int u = q[front];
front++;
if (u == n+1)
return true;
for (int i=head[u]; i!=-1; i=edge[i].next) {
int v = edge[i].t;
if (level[v] == -1 && edge[i].v) {
level[v] = level[u]+1;
q[tail++] = v;;
}
}
}
return false;
}

int dfs(int u, int minf) {  // 增广多条路
if (u == n+1)
return minf;
int ret  = 0;   //记录所有增广路的流量和
for (int i=head[u]; i!=-1; i=edge[i].next) {
int v = edge[i].t, f;
if (level[v] == level[u]+1 && edge[i].v) {
int Min = min(minf-ret, edge[i].v);
f = dfs(v, Min);
edge[i].v -= f;
edge[i^1].v += f;
ret += f;
if (ret == minf)
return ret;
}
}
return ret;
}

int dinic() {
int flow = 0;
while (bfs())
flow += dfs(0, inf);
return flow;
}

int main()
{
while (scanf("%d%d%d", &K, &C, &M)!=EOF) {
n = K+C;
for (int i=1; i<=n; i++) {
for (int j=1; j<=n; j++)
{
scanf("%d", &dis[i][j]);
if (dis[i][j] == 0)
dis[i][j] = inf;
}
}
for (int k=1; k<=n; k++) {
for (int i=1; i<=n; i++) {
for (int j=1; j<=n; j++)
dis[i][j] = min(dis[i][j], dis[i][k]+dis[k][j]);
}
}
int l=0, r=300*n;
int ans=0;
while (r-l > 1) {
int mid = (l+r)/2;
build(mid);
int F;
F = dinic();
if (F >= C) {
r = mid;
ans = mid;
}
else
l = mid;
}
printf("%d\n", ans);
}
return 0;
}