BZOJ 1189 HNOI 2007 紧急疏散 evacuate 二分答案 最大流

标签：bzoj   bzoj1189   二分答案   网络流   最大流   

题目大意：紧急疏散。有一张地图，‘.’表示人，‘D’表示门，人需要走曼哈顿距离的单位时间才1能到达门。一个门一个时刻只能通过一个人。求多长时间能疏散完毕。

思路：二分答案＋最大流满流判定。先BFS处理出每个人与门的距离。二分最小时间，然后连边。S向每个人连流量为1的边，每个人向二分的时间之内能到达的门连流量为1的边。每个门向T连流量为t的边。然后最大流判定是否满流。

（数组大小我是瞎开的，写代码的时候要算好了在开！）

CODE：

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 10010
#define S 0
#define T 5000
#define INF 0x7f7f7f7f
using namespace std;
const int dx[] = {0,0,0,1,-1};
const int dy[] = {0,1,-1,0,0};

struct Door{
	int x,y;
	Door(int _,int __):x(_),y(__) {}
	Door() {}
}door[MAX];
struct Empty{
	int x,y;
	Empty(int _,int __):x(_),y(__) {}
	Empty() {}
}person[MAX];
struct Status{
	int x,y;
	int length;

	Status(int _,int __,int ___)
		:x(_),y(__),length(___) {}
	Status() {}
};

int m,n;
int doors,persons;
char src[30][30];
int num[30][30],dis[MAX][410];
bool v[30][30];

int head[MAX],total;
int _next[2000000],aim[2000000],flow[2000000];

int deep[MAX];

inline void BFS(int p);
inline void MakeGraph(int t);
inline void Initialize();
inline void Add(int x,int y,int f);

bool BFS();
int Dinic(int x,int f);

int main()
{
	cin >> m >> n;
	for(int i = 1;i <= m; ++i) {
		scanf("%s",src[i] + 1);
		for(int j = 1;j <= n; ++j) {
			if(src[i][j] == 'D') {
				door[++doors] = Door(i,j);
				num[i][j] = doors;
			}
			else if(src[i][j] == '.') {
				person[++persons] = Empty(i,j);
				num[i][j] = persons;
			}
		}
	}
    memset(dis,0x3f,sizeof(dis));
	for(int i = 1;i <= doors; ++i)
		BFS(i);
	int l = 0,r = 400,ans = -1;
	while(l <= r) {
		int mid = (l + r) >> 1;
		MakeGraph(mid);
		int max_flow = 0;
		while(BFS())
			max_flow += Dinic(S,INF);
		if(max_flow == persons)
			ans = mid,r = mid - 1;
		else	l = mid + 1;
	}
	if(ans == -1)	puts("impossible");
	else	cout << ans << endl;
	return 0;
}

inline void BFS(int p)
{
	static queue<Status> q;
    while(!q.empty())   q.pop();
	memset(v,false,sizeof(v));
	q.push(Status(door[p].x,door[p].y,0));
	while(!q.empty()) {
		Status now = q.front(); q.pop();
		v[now.x][now.y] = true;
		for(int i = 1;i <= 4; ++i) {
			int fx = now.x + dx[i];
			int fy = now.y + dy[i];
			if(v[fx][fy] || !fx || !fy || fx > m || fy > n)	continue;
            if(src[fx][fy] == '.') {
                dis[num[fx][fy]][p] = now.length + 1;
                q.push(Status(fx,fy,now.length + 1));
            }
		}
	}
}

inline void MakeGraph(int t)
{
	Initialize();
	for(int i = 1;i <= persons; ++i)
		Add(S,i,1),Add(i,S,0);
	for(int i = persons + 1;i <= persons + doors; ++i)
		Add(i,T,t),Add(T,i,0);
	for(int i = 1;i <= persons; ++i)
		for(int j = 1;j <= doors; ++j)
			if(dis[i][j] <= t)
				Add(i,j + persons,1),Add(j + persons,i,0);
}

inline void Initialize()
{
	total = 1;
	memset(head,0,sizeof(head));
}

inline void Add(int x,int y,int f)
{
	_next[++total] = head[x];
	aim[total] = y;
	flow[total] = f;
    head[x] = total;
}

bool BFS()
{
	static queue<int> q;
    while(!q.empty())   q.pop();
	memset(deep,0,sizeof(deep));
	deep[S] = 1;
	q.push(S);
	while(!q.empty()) {
		int x = q.front(); q.pop();
		for(int i = head[x];i;i = _next[i])
			if(flow[i] && !deep[aim[i]]) {
				deep[aim[i]] = deep[x] + 1;
				q.push(aim[i]);
				if(aim[i] == T)	return true;
			}
	}
	return false;
}

int Dinic(int x,int f)
{
	if(x == T)	return f;
	int temp = f;
	for(int i = head[x];i;i = _next[i])
		if(flow[i] && temp && deep[aim[i]] == deep[x] + 1) {
			int away = Dinic(aim[i],min(flow[i],temp));
			flow[i] -= away;
			flow[i^1] += away;
			temp -= away;
		}
    if(temp == f)   deep[x] = 0;
	return f - temp;
}

BZOJ 1189 HNOI 2007 紧急疏散 evacuate 二分答案 最大流

标签：bzoj   bzoj1189   二分答案   网络流   最大流   

原文地址：http://blog.csdn.net/jiangyuze831/article/details/39737175