poj_2112 网络最大流+二分法

题目大意

    有K台挤奶机和C头奶牛，都被视为物体，这K+C个物体之间存在路径。给出一个 (K+C)x(K+C) 的矩阵A，A[i][j]表示物体i和物体j之间的距离，有些物体之间可能没有直接通路。 
    每台挤奶机可以容纳m头奶牛去挤奶，且每个奶牛仅可以去往一台挤奶机。现在安排这C头奶牛去挤奶，每头奶牛会去往某个挤奶机，求出这C头奶牛去其对应挤奶机的路径长度的最大值的最小值。

题目分析

    “每头奶牛仅可以去往一台挤奶机，每台挤奶机最多有M头奶牛”这似乎是一个路径流量的问题，考虑使用网络流算法来解决。 
    那么，如何确定最长路径的最小值呢？可以先考虑简化问题，“给定一个最大距离L，能否分配这C头奶牛的挤奶机，使得每头奶牛到达挤奶机的距离都小于L”。 
    解决简化问题：虚拟一个源点和汇点。从源点引出C条容量分别为1的路径到达C头奶牛，再将K台机器分别引出一条容量为M的路径到达汇点。则问题转化为，构造C头奶牛到K台机器的网络路径，且从每头奶牛去往挤奶机的路径的容量为1，距离不超过L，使得网络从源点到汇点的最大流量为C，且C头奶牛走的路径的最大值最小。 
    然后，再通过二分法枚举最大距离L，找到最大距离的最小值。 
    具体实现的时候，使用Floyd算法求出奶牛到达挤奶机的最短路径长度；使用ISAP算法找出最大流；使用二分法确定最长的路径最小值。

实现(c++)

#include<stdio.h>
#include<string.h>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
#define MAX_NODE 235
#define MAX_EDGE_NUM 50000
#define INFINITE 1 << 20
#define min(a, b) a < b?a:b
struct Edge{
	int from;
	int to;
	int w;
	int next;
	int rev;
	bool operator==(const pair<int, int>& p){
		return from == p.first && to == p.second;
	}
};

Edge gEdges[MAX_EDGE_NUM];
int gFlow[MAX_NODE][MAX_NODE];
int gHead[MAX_NODE];
int gDist[MAX_NODE];
int gGap[MAX_NODE];
int gPre[MAX_NODE];
int gPath[MAX_NODE];

int gEdgeCount;
int gSource, gDestination;
void InsertEdge(int u, int v, int w){
	Edge* it = find(gEdges, gEdges + gEdgeCount, pair<int, int>(u, v));
	if (it != gEdges + gEdgeCount){
		it->w = w;
	}
	else{
		int e1 = gEdgeCount++;
		gEdges[e1].from = u;
		gEdges[e1].to = v;
		gEdges[e1].w = w;
		gEdges[e1].next = gHead[u];
		gHead[u] = e1;

		int e2 = gEdgeCount++;
		gEdges[e2].from = v;
		gEdges[e2].to = u;
		gEdges[e2].w = 0;
		gEdges[e2].next = gHead[v];
		gHead[v] = e2;

		gEdges[e1].rev = e2;
		gEdges[e2].rev = e1;
	}
	gFlow[u][v] = w;
}

void Bfs(){
	memset(gGap, 0, sizeof(gGap));
	memset(gDist, -1, sizeof(gDist));
	gDist[gDestination] = 0;
	gGap[0] = 1;
	queue<int>Q;
	Q.push(gDestination);
	while (!Q.empty()){
		int u = Q.front();
		Q.pop();
		for (int e = gHead[u]; e != -1; e = gEdges[e].next){
			int v = gEdges[e].to;
			if (gDist[v] >= 0)
				continue;
			gDist[v] = gDist[u] + 1;
			gGap[gDist[v]] ++;
			Q.push(v);
		}
	}
}

int ISAP(int n){
	int e, d, u = gSource;
	int ans = 0;
	Bfs();
	while (gDist[gSource] <= n){
		if (u == gDestination){
			int min_flow = INFINITE;
			for (e = gPath[u]; u != gSource; e = gPath[u = gPre[u]]){
				min_flow = min(min_flow, gEdges[e].w);
			}
			u = gDestination;
			for (e = gPath[u]; u != gSource; e = gPath[u = gPre[u]]){
				gEdges[e].w -= min_flow;
				gEdges[gEdges[e].rev].w += min_flow;

				gFlow[gPre[u]][u] -= min_flow;
				gFlow[u][gPre[u]] += min_flow;
			}
			ans += min_flow;
		}
		for (e = gHead[u]; e != -1; e = gEdges[e].next){
			if (gEdges[e].w > 0 && gDist[u] == gDist[gEdges[e].to] + 1)
				break;
		}
		if (e >= 0){
			gPre[gEdges[e].to] = u;
			gPath[gEdges[e].to] = e;
			u = gEdges[e].to;
		}
		else{
			if (--gGap[gDist[u]] == 0)
				break;
			d = n;
			for (int e = gHead[u]; e != -1; e = gEdges[e].next)
				if (gEdges[e].w > 0)
					d = min(d, gDist[gEdges[e].to]);
			gDist[u] = d + 1;
			++gGap[gDist[u]];
			if (u != gSource)
				u = gPre[u];
		}
	}
	return ans;
}

int gMinDist[MAX_NODE][MAX_NODE];
void Floyd(int n){
	for (int k = 1; k <= n; k++){
		for (int i = 1; i <= n; i++){
			for (int j = 1; j <= n; j++){
				if (gMinDist[i][j] > gMinDist[i][k] + gMinDist[k][j]){
					gMinDist[i][j] = gMinDist[i][k] + gMinDist[k][j];
				}
			}
		}
	}
}

void BuildGraph(int k, int c, int m, int max_dist){
	memset(gHead, -1, sizeof(gHead));
	gEdgeCount = 0;
	gSource = 0;
	gDestination = k + c + 1;
	
	for (int i = k + 1; i <= k + c; i++){
		InsertEdge(gSource, i, 1);
	}
	for (int i = k + 1; i <= k + c; i++){ 
		for (int j = 1; j <= k; j++){
			if (gMinDist[i][j] <= max_dist)
				InsertEdge(i, j, 1);
		}		
	}

	for (int i = 1; i <= k; i++){
		InsertEdge(i, gDestination, m);
	}
}
void print_graph(int n){
	for (int u = 0; u <= n; u++){
		printf("node %d links to ", u);
		for (int e = gHead[u]; e != -1; e = gEdges[e].next)
			printf("%d(flow = %d) ", gEdges[e].to, gEdges[e].w);
		printf("
");
	}
}
int main(){
	int k, c, m, d;
	while (scanf("%d %d %d", &k, &c, &m) != EOF){
		for (int i = 1; i <= k + c; i++){
			for (int j = 1; j <= k + c; j++){
				scanf("%d", &gMinDist[i][j]);
				if (i != j && gMinDist[i][j] == 0)
					gMinDist[i][j] = INFINITE;
			}
		}

		Floyd(k + c);
		int  beg = 1, end = 40010;
		while (beg < end){
			int max_dist = (beg + end) / 2;
			BuildGraph(k, c, m, max_dist);
			//	print_graph(k + c + 1);
			int max_flow = ISAP(k + c + 2);
			if (max_flow == c)
				end = max_dist;
			else
				beg = max_dist + 1;
		}
		printf("%d
", end);
	}
	return 0;
}