Poj2195 going home [minimum cost flow] + [optimal matching of Bipartite Graph]

Going home

Time limit:1000 ms		Memory limit:65536 K
Total submissions:18169		Accepted:9268

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 done 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, '. '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, aggressive. 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.mH.5 5HH..m...............mm..H7 8...H.......H.......H....mmmHmmmm...H.......H.......H....0 0

Sample output

21028

Source

Pacific Northwest 2004: given an N * m graph '. 'The open space, the villain can go, 'M' is the villain, 'H' is the house, and the villain can pass by. The villain can only walk along the upper, lower, and lower sides of a grid at a time. Now, every villain is required to enter a different house, and the minimum number of steps is required. Problem: this problem can be solved as the minimum charge flow. The distance from the villain to the House is the cost, the capacity is 1, and the capacity from the source point to each villain is set to 1, and the cost is 0, the cost of each house to the settlement point is 0, and the capacity is 1. The rest is to find the minimum fee flow. Version 1: Minimum fee flow: 125 Ms

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <queue>const int maxn = 205;const int inf = 0x3f3f3f3f;char str[maxn];int head[maxn], n, m; // n rows, m columnsstruct Node {    int x, y;} A[maxn], B[maxn];int id1, id2, id, source, sink;struct node {    int f, c, u, v, next;} E[maxn * maxn];bool vis[maxn];int pre[maxn], dist[maxn];void addEdge(int u, int v, int c, int f) {    E[id].u = u; E[id].v = v;    E[id].c = c; E[id].f = f;    E[id].next = head[u]; head[u] = id++;    E[id].u = v; E[id].v = u;    E[id].c = 0; E[id].f = -f;    E[id].next = head[v]; head[v] = id++;}void getMap() {    int i, j, dis; Node e;     id = id1 = id2 = 0;    for(i = 0; i < n; ++i) {        scanf("%s", str);        for(j = 0; str[j] != '\0'; ++j) {            if(str[j] == '.') continue;            e.x = i; e.y = j;            if(str[j] == 'm') A[id1++] = e;            else B[id2++] = e;        }    }    memset(head, -1, sizeof(head));    source = id1 + id2; sink = source + 1;    for(i = 0; i < id1; ++i) {        for(j = 0; j < id2; ++j) {            dis = abs(A[i].x - B[j].x) + abs(A[i].y - B[j].y);            addEdge(i, id1 + j, 1, dis); // uvcf        }        addEdge(source, i, 1, 0);    }    for(j = 0; j < id2; ++j)        addEdge(id1 + j, sink, 1, 0);}bool SPFA(int start, int end) {    std::queue<int> Q; int i, u, v;    memset(vis, 0, sizeof(vis));    memset(pre, -1, sizeof(pre));    memset(dist, 0x3f, sizeof(pre));    Q.push(start); vis[start] = 1; dist[start] = 0;    while(!Q.empty()) {        u = Q.front(); Q.pop();        vis[u] = 0;        for(i = head[u]; i != -1; i = E[i].next) {            v = E[i].v;            if(E[i].c && dist[v] > dist[u] + E[i].f) {                dist[v] = dist[u] + E[i].f;                pre[v] = i;                if(!vis[v]) {                    Q.push(v); vis[v] = 1;                }            }        }    }    return dist[end] != inf;}int Min_Cost_Flow(int start, int end) {    int ans_cost = 0, u, minCut;    while(SPFA(start, end)) {        minCut = inf;        for(u = pre[end]; u != -1; u = pre[E[u].u]) {            if(minCut > E[u].c) minCut = E[u].c;        }        for(u = pre[end]; u != -1; u = pre[E[u].u]) {            E[u].c -= minCut; E[u^1].c += minCut;        }        ans_cost += minCut * dist[end];    }    return ans_cost;}void solve() {    printf("%d\n", Min_Cost_Flow(source, sink));}int main() {    // freopen("stdin.txt", "r", stdin);    while(scanf("%d%d", &n, &m), n | m) {        getMap();        solve();    }    return 0;}

Version 2: km: 0 ms

#include <stdio.h>#include <string.h>#include <stdlib.h>const int maxn = 105;const int largeNum = 210;const int inf = 0x3f3f3f3f;int n, m; // n rows, m columnschar str[maxn];struct Node {    int x, y;} A[maxn], B[maxn];int id1, id2;int G[maxn][maxn];int Lx[maxn], Ly[maxn];int match[maxn];bool visx[maxn], visy[maxn];int slack[maxn];void getMap() {    int i, j, dis; Node e;    id1 = id2 = 0;    for(i = 0; i < n; ++i) {        scanf("%s", str);        for(j = 0; str[j] != '\0'; ++j) {            if(str[j] == '.') continue;            e.x = i; e.y = j;            if(str[j] == 'm') A[id1++] = e;            else B[id2++] = e;        }    }    memset(G, 0, sizeof(G));    for(i = 0; i < id1; ++i) {        for(j = 0; j < id2; ++j) {            G[i][j] = largeNum - (abs(A[i].x - B[j].x) + abs(A[i].y - B[j].y));        }    }}bool DFS(int cur) {    int t, y;    visx[cur] = true;    for(y = 0; y < id2; ++y) {        if(visy[y]) continue;        t = Lx[cur] + Ly[y] - G[cur][y];        if(t == 0) {            visy[y] = true;            if(match[y] == -1 || DFS(match[y])) {                match[y] = cur; return true;            }        } else if(slack[y] > t) slack[y] = t;     }    return false;}int KM() {    int i, j, x, d, ret;    memset(match, -1, sizeof(match));    memset(Ly, 0, sizeof(Ly));    for(i = 0; i < id1; ++i) {        Lx[i] = -inf;        for(j = 0; j < id2; ++j)            if(G[i][j] > Lx[i]) Lx[i] = G[i][j];    }    for(x = 0; x < id1; ++x) {        memset(slack, 0x3f, sizeof(slack));        while(true) {            memset(visx, 0, sizeof(visx));            memset(visy, 0, sizeof(visy));            if(DFS(x)) break;            d = inf;            for(i = 0; i < id2; ++i)                if(!visy[i] && d > slack[i])                     d = slack[i];            for(i = 0; i < id1; ++i)                if(visx[i]) Lx[i] -= d;            for(i = 0; i < id2; ++i)                if(visy[i]) Ly[i] += d;                else slack[i] -= d;        }    }    ret = 0;    for(i = 0; i < id1; ++i)        if(match[i] > -1) ret += G[match[i]][i];    return ret;}void solve() {    printf("%d\n", largeNum * id1 - KM());}int main() {    // freopen("stdin.txt", "r", stdin);    while(scanf("%d%d", &n, &m), n | m) {        getMap();        solve();    }    return 0;}

Poj2195 going home [minimum cost flow] + [optimal matching of Bipartite Graph]