HDOJ：1533-Going Home（最小费用流）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1533

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解题心得：

• 第一次写最小费用流的题，去hdoj上找了一个入门级题目，建图比较简单，用了spfa和Dij两种写法。

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//spfa
#include <bits/stdc++.h>
using namespace std;
const int maxn = 110*110;

struct edge{
    int to, cap, dis, rev;

    edge(int To, int Cap, int Dis, int Rev):
            to(To), cap(Cap), dis(Dis), rev(Rev) {}
};

//感觉这里写得超级智障，懒得改了
struct men {
    int x,y;
};

struct House {
    int x,y;
};

vector <edge> ve[maxn];
vector <men> man;
vector <House> house;
int n, m, S, T;
char maps[maxn][maxn];


void init() {
    S = 0;
    man.clear();
    house.clear();
    for(int i=0;i<maxn;i++)
        ve[i].clear();
    for(int i=1;i<=n;i++)
        scanf("%s",maps[i]+1);
    for(int i=1;i<=n;i++) {
        for(int j=1;j<=m;j++) {
            if(maps[i][j] == 'm') {
                men now;
                now.x = i;
                now.y = j;
                man.push_back(now);
            } else if(maps[i][j] == 'H') {
                House temp;
                temp.x = i;
                temp.y = j;
                house.push_back(temp);
            }
        }
    }
    T = man.size() + house.size() + 1;
}

void build_adge(int s, int to, int dis) {
    ve[s].push_back(edge(to, 1, dis, ve[to].size()));
    ve[to].push_back(edge(s, 0, -dis, ve[s].size()-1));
}

void add_adge() {
    for(int i=0;i<man.size();i++) {
        build_adge(S, i+1, 0);
        for(int j=0;j<house.size();j++) {
            int dis = abs(man[i].x - house[j].x) + abs(man[i].y - house[j].y);
            build_adge(i+1, man.size()+j+1, dis);
        }
    }
    for(int i=0;i<house.size();i++) {
        build_adge(i+1+man.size(), T, 0);
    }
}

int dist[maxn], preve[maxn], prevv[maxn];
bool vis[maxn];
bool SPFA() {
    memset(preve, 0, sizeof(preve));
    memset(vis, 0, sizeof(vis));
    memset(prevv, 0, sizeof(prevv));
    memset(dist, 0x3f, sizeof(dist));
    dist[S] = 0;
    queue <int> qu;
    qu.push(S);
    while(!qu.empty()) {
        int now = qu.front() ;qu.pop();
        vis[now] = false;
        for(int i=0;i<ve[now].size();i++) {
            edge &e = ve[now][i];
            if(dist[e.to] > dist[now] + e.dis && e.cap > 0) {
                dist[e.to] = dist[now] + e.dis;
                prevv[e.to] = now;
                preve[e.to] = i;
                if(!vis[e.to]) {
                    qu.push(e.to);
                    vis[e.to] = true;
                }
            }
        }
    }
    return dist[T] != 0x3f3f3f3f;
}

int min_cost_flow() {
    int ans = 0;
    while(SPFA()) {
        for(int v = T; v!=S ; v =prevv[v]) {
            edge &e = ve[prevv[v]][preve[v]];
            e.cap -= 1;
            ve[v][e.rev].cap += 1;
            ans += e.dis;
        }
    }
    return ans;
}

int main() {
    while(scanf("%d%d",&n,&m) && n+m) {
        init();
        add_adge();
        printf("%d
",min_cost_flow());
    }
    return 0;
}

---

Dij

#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> P;
const int maxn = 110*110;

struct edge{
    int to, cap, dis, rev;

    edge(int To, int Cap, int Dis, int Rev):
            to(To), cap(Cap), dis(Dis), rev(Rev) {}
};

struct men {
    int x,y;
};

struct House {
    int x,y;
};

vector <edge> ve[maxn];
vector <men> man;
vector <House> house;
int n, m, S, T, h[maxn];
char maps[maxn][maxn];


void init() {
    S = 0;
    man.clear();
    house.clear();
    memset(h, 0, sizeof(h));
    for(int i=0;i<maxn;i++)
        ve[i].clear();
    for(int i=1;i<=n;i++)
        scanf("%s",maps[i]+1);
    for(int i=1;i<=n;i++) {
        for(int j=1;j<=m;j++) {
            if(maps[i][j] == 'm') {
                men now;
                now.x = i;
                now.y = j;
                man.push_back(now);
            } else if(maps[i][j] == 'H') {
                House temp;
                temp.x = i;
                temp.y = j;
                house.push_back(temp);
            }
        }
    }
    T = man.size() + house.size() + 1;
}

void build_adge(int s, int to, int dis) {
    ve[s].push_back(edge(to, 1, dis, ve[to].size()));
    ve[to].push_back(edge(s, 0, -dis, ve[s].size()-1));
}

void add_adge() {
    for(int i=0;i<man.size();i++) {
        build_adge(S, i+1, 0);
        for(int j=0;j<house.size();j++) {
            int dis = abs(man[i].x - house[j].x) + abs(man[i].y - house[j].y);
            build_adge(i+1, man.size()+j+1, dis);
        }
    }
    for(int i=0;i<house.size();i++) {
        build_adge(i+1+man.size(), T, 0);
    }
}

int dist[maxn], preve[maxn], prevv[maxn];
bool Dij() {
    memset(preve, 0, sizeof(preve));
    memset(prevv, 0, sizeof(prevv));
    memset(dist, 0x3f, sizeof(dist));
    dist[S] = 0;
    priority_queue <P, vector<P>, greater<P> > qu;
    qu.push(make_pair(0,0));
    while(!qu.empty()) {
        P now = qu.top(); qu.pop();
        int pos = now.second;
        if(dist[pos] < now.first)
            continue;
        for(int i=0;i<ve[pos].size();i++) {
            edge &e = ve[pos][i];
            if(e.cap > 0 && dist[e.to] > dist[pos] + e.dis + h[pos] - h[e.to]) {
                dist[e.to] = dist[pos] + e.dis + h[pos] - h[e.to];
                prevv[e.to] = pos;
                preve[e.to] = i;
                qu.push(make_pair(dist[e.to], e.to));
            }
        }
    }
    return dist[T] != 0x3f3f3f3f;
}

int min_cost_flow() {
    int ans = 0;
    while(Dij()) {
        for(int i=0;i<=T;i++)
            h[i] += dist[i];
        ans += h[T];
        for(int v = T; v!=S ; v =prevv[v]) {
            edge &e = ve[prevv[v]][preve[v]];
            e.cap -= 1;
            ve[v][e.rev].cap += 1;
        }
    }
    return ans;
}

int main() {
    while(scanf("%d%d",&n,&m) && n+m) {
        init();
        add_adge();
        printf("%d
",min_cost_flow());
    }
    return 0;
}