HDU Problem - 1533 Going Home（费用流板子题）

题目链接

Problem Description

On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertically, to an adjacent point. For each little man, you need to pay a $1 travel fee for every step he moves, until he enters a house. The task is complicated with the restriction that each house can accommodate only one little man. Your task is to compute the minimum amount of money you need to pay in order to send these n little men into those n different houses. The input is a map of the scenario, a ‘.’ means an empty space, an ‘H’ represents a house on that point, and am ‘m’ indicates there is a little man on that point. You can think of each point on the grid map as a quite large square, so it can hold n little men at the same time; also, it is okay if a little man steps on a grid with a house without entering that house.

Input

There are one or more test cases in the input. Each case starts with a line giving two integers N and M, where N is the number of rows of the map, and M is the number of columns. The rest of the input will be N lines describing the map. You may assume both N and M are between 2 and 100, inclusive. There will be the same number of ‘H’s and ‘m’s on the map; and there will be at most 100 houses. Input will terminate with 0 0 for N and M.

Output

For each test case, output one line with the single integer, which is the minimum amount, in dollars, you need to pay.

Sample Input

2 2
.m
H.
5 5
HH..m
.....
.....
.....
mm..H
7 8
...H....
...H....
...H....
mmmHmmmm
...H....
...H....
...H....
0 0

Sample Output

2
10
28

AC

建边
1. 源点到people建边，流量为1，费用为0
2. 所有的people到所有的house建边，流量为1，费用是曼哈顿距离
3. house到汇点建边，流量为1，费用为0
费用流模板

#include <iostream>
#include <stdio.h>
#include <map>
#include <vector>
#include <queue>
#include <cmath>
#include <algorithm>
#include <cmath>
#define N 80005
#include <cstring>
#define ll long long
#define P pair<int, int>
#define mk make_pair
using namespace std;
// 本来是prev记录前驱节点的标号，但是hdu不让用，这里改为fuck
int fuck[10200], dis[10200], head[10200] ;
bool vis[10200];
int n, m, cnt;
int inf = 0x3f3f3f3f;struct ac{int v, c, cost, pre;
}edge[N];void addedge(int u, int v, int c, int cost) {edge[cnt].v = v;edge[cnt].c = c;edge[cnt].cost = cost;edge[cnt].pre = head[u];head[u] = cnt++;swap(u, v);edge[cnt].v = v;edge[cnt].c = 0;edge[cnt].cost = -cost;edge[cnt].pre = head[u];head[u] = cnt++;
}
bool spfa(int s, int e) {memset(vis, false, sizeof(vis));memset(dis, inf, sizeof(dis));memset(fuck, -1, sizeof(fuck));queue<int> que;que.push(s);dis[s] = 0;vis[s] = true;while (!que.empty()) {int u = que.front();que.pop();vis[u] = false;for (int i = head[u]; i != -1; i = edge[i].pre) {int v = edge[i].v;int c = edge[i].c;int cost = edge[i].cost;if (dis[v] > dis[u] + cost && c > 0) {dis[v] = dis[u] + cost;fuck[v] = i;if (!vis[v]) {vis[v] = true;que.push(v);}}}}if (dis[e] == inf)return false;elsereturn true;
}int mincost(int s, int e, int &cost) {int maxflow = 0;int minflow = inf;while (spfa(s, e)) {for (int i = fuck[e]; i != -1; i = fuck[edge[i ^ 1].v])minflow = min(minflow, edge[i].c);for (int i = fuck[e]; i != -1; i = fuck[edge[i ^ 1].v]) {edge[i].c -= minflow;edge[i ^ 1].c += minflow;cost += minflow * edge[i].cost;}maxflow += minflow;}return maxflow;
}
// 存放hose, people信息
struct cell {int num, x, y;
};
void debug(int a, int b, int c, int cost) {
//  cout << a << " -> " << b <<  " " << c << " " << cost << endl;
}
char g[102][102];
int main() {
#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);
#endifwhile (scanf("%d%d", &n, &m), n + m) {memset(head, -1, sizeof(head));cnt = 0;int start = 0, end = n * m + 1;for (int i = 0; i < n; ++i) {scanf("%s", g[i]);}// 统计 people和house 的信息 vector<cell> house, people;for (int i = 0; i < n; ++i) {for (int j = 0; j < m; ++j) {if (g[i][j] == '.') continue;if (g[i][j] == 'H') {int num = i * m + j + 1;people.push_back((cell){num, i, j});}if (g[i][j] == 'm') {int num = i * m + j + 1;house.push_back((cell){num, i, j});}}}int house_num = house.size();int people_num = people.size();// 源点到people建边，花费0， 流量1 for (int i = 0; i < people_num; ++i) {cell temp_people = people[i];addedge(start, temp_people.num, 1, 0);debug(start, temp_people.num, 1, 0);}// house到汇点建边， 花费0， 流量1 for (int i = 0; i < house_num; ++i) {cell temp_house = house[i];addedge(temp_house.num, end, 1, 0);debug(temp_house.num, end, 1, 0);}// people到house建边，花费为曼哈顿距离，流量为1 for (int i = 0; i < people_num; ++i) {for (int j = 0; j < house_num; ++j) {cell temp_house = house[j];cell temp_people = people[i];int temp_dis = fabs(temp_house.x - temp_people.x) + fabs(temp_house.y - temp_people.y);addedge(temp_people.num, temp_house.num, 1, temp_dis);debug(temp_people.num, temp_house.num, 1, temp_dis);}}int ans = 0;        mincost(start, end, ans);printf("%d\n", ans);}return 0;
}