最小费用最大流 I

hdu 2485 Destroying the bus stations

第一次接触这类问题，最小费用最大流的关键是如何建图，其次就是套模板。

这题借鉴了别人的方法：由于图中的两个节点间可能互有通路，所以将一个点拆分成两个，分别表示入和出。绝大多数模板中都是不断求源端到汇点的最小费用路径然后扩增。因此，将每次扩增等于删除流量最小段所代表的点。直到源点到汇点的最小费用大于k

#include<cstdio>
#include<algorithm>
#include<vector>
#include<cmath>
#include<set>
#pragma comment(linker, "/STACK:1024000000,1024000000")
// acm.hdu.edu.cn/showproblem.php?pid=4494
using namespace std;
#define MAXN 105
int n, m, k;
struct _node
{
    int to, next, cost, val;
    _node(int a, int b, int c, int d): to(a),next(b),cost(c),val(d) {}
    _node() {}
}ee[MAXN*MAXN];
int head[MAXN], cnt;
int end ;
void add(int u, int v, int cost, int val)
{
    ee[cnt] = _node(v, head[u], cost, val);
    head[u] = cnt++;
    ee[cnt] = _node(u, head[v], -cost, 0);
    head[v] = cnt++;
}
int q[MAXN*MAXN], vis[MAXN], dis[MAXN], fa[MAXN];
int pos[MAXN];
int spfa()
{
    int bg = 0, ed = 0;
    memset(vis, 0, sizeof vis);
    memset(fa, -1, sizeof fa);
    memset(dis, 0x3f, sizeof dis);
    dis[1] = 0;
    q[ed++] = 1;
    while (bg < ed)
    {
        int u = q[bg++];
        vis[u] = 0;
        for (int i = head[u]; ~i; i=ee[i].next)
        {
            int v = ee[i].to;
            if (ee[i].val > 0 && dis[v]>dis[u]+ee[i].cost)
            {
                dis[v]=dis[u]+ee[i].cost;
                fa[v] = u;
                pos[v] = i;
                if (!vis[v])
                {
                    vis[v] = 1;
                    q[ed++] = v;
                }
            }
        }
    }
    if (dis[end] > k) return 0;
    return fa[end] != -1;
}
int minCost_maxFlow()
{
    int flow = 0;
    int res = 0;
    while (spfa())
    {
        int mn = 0x3fffffff;
        for (int i = end; i!= 1; i = fa[i])
        {
            if (mn > ee[pos[i]].val)
                mn = ee[pos[i]].val;
        }
        for (int i = end; i!=1; i=fa[i])
        {
            ee[pos[i]].val -= mn;
            ee[pos[i]^1].val += mn;
        }
        flow += mn;
        res += dis[end];
    }
    return flow;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif
    while(scanf("%d%d%d", &n, &m, &k) && (n+m+k))
    {
        end = n;
        cnt = 0;
        memset(head, -1, sizeof head);
        for (int i = 2; i< n; ++i)
            add(i, i+n, 0, 1);
        for (int i = 0; i< m; ++i)
        {
            int u, v;
            scanf("%d%d", &u, &v);
            if (u == 1)
                add(u, v, 1, 1);
            else
                add(u+n,v,1, 1);
        }
        printf("%d\n", minCost_maxFlow());    
    }
    return 0;
}

hdu 1853 Cyclic Tour

若图中所有的点以圆环的形式存在，当且仅当每个点入度和出度均为1

#include<cstdio>
#include<algorithm>
#include<vector>
#include<cmath>
#include<set>
#pragma comment(linker, "/STACK:1024000000,1024000000")
// acm.hdu.edu.cn/showproblem.php?pid=1853
using namespace std;
#define MAXN 205
int n, m;
struct _node
{
    int to, next, cost, val;
    _node(int a, int b, int c, int d): to(a),next(b),cost(c),val(d) {}
    _node() {}
}ee[MAXN*MAXN];
int head[MAXN], cnt;
int end ;
void add(int u, int v, int cost, int val)
{
    ee[cnt] = _node(v, head[u], cost, val);
    head[u] = cnt++;
    ee[cnt] = _node(u, head[v], -cost, 0);
    head[v] = cnt++;
}
int q[MAXN*MAXN], vis[MAXN], dis[MAXN], fa[MAXN];
int pos[MAXN];
int spfa()
{
    int bg = 0, ed = 0;
    memset(vis, 0, sizeof vis);
    memset(fa, -1, sizeof fa);
    memset(dis, 0x3f, sizeof dis);
    dis[0] = 0;
    q[ed++] = 0;
    while (bg < ed)
    {
        int u = q[bg++];
        vis[u] = 0;
        for (int i = head[u]; ~i; i=ee[i].next)
        {
            int v = ee[i].to;
            if (ee[i].val > 0 && dis[v]>dis[u]+ee[i].cost)
            {
                dis[v]=dis[u]+ee[i].cost;
                fa[v] = u;
                pos[v] = i;
                if (!vis[v])
                {
                    vis[v] = 1;
                    q[ed++] = v;
                }
            }
        }
    }
    return fa[end] != -1;
}
int minCost_maxFlow()
{
    int flow = 0;
    int res = 0;
    while (spfa())
    {
        int mn = 0x3fffffff;
        for (int i = end; i!= 0; i = fa[i])
        {
            if (mn > ee[pos[i]].val)
                mn = ee[pos[i]].val;
        }
        for (int i = end; i!=0; i=fa[i])
        {
            ee[pos[i]].val -= mn;
            ee[pos[i]^1].val += mn;
        }
        flow += mn;
        res += dis[end];
    }
    return flow!=n?-1:res;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif
    while(scanf("%d%d", &n, &m) !=EOF)
    {
        end = 2*n+1;
        cnt = 0;
        memset(head, -1, sizeof head);
        for (int i = 1; i<= n; ++i)
            add(0,i,0,1);
        for (int i = 0; i< m; ++i)
        {
            int u, v, c;
            scanf("%d%d%d", &u, &v, &c);
            add(u, v+n, c, 1);
        }
        for (int i = n+1; i< end; ++i)
            add(i, end, 0, 1);    
        printf("%d\n", minCost_maxFlow());    
    }
    return 0;
}

hdu 4494 Teamwork

主要参考了 http://www.cnblogs.com/wangfang20/p/3280172.html 该博客。

各个种类间是不会互相影响的。对于有向图一般把一个点拆分成入和出两点，但题中由于工人可复用，所以再次引入一个点表示重复利用的工人。

#include<cstdio>
#include<algorithm>
#include<vector>
#include<cmath>
#include<set>
#pragma comment(linker, "/STACK:1024000000,1024000000")
// acm.hdu.edu.cn/showproblem.php?pid=4494
using namespace std;
const int MAXN = 155*3;
int n, m;
struct _node
{
	int to, next, cost, val;
	_node(int a, int b, int c, int d): to(a),next(b),cost(c),val(d) {}
	_node() {}
}ee[MAXN*200];
int head[MAXN], cnt;
void add(int u, int v, int cost, int val)
{
	ee[cnt] = _node(v, head[u], cost, val);
	head[u] = cnt++;
	ee[cnt] = _node(u, head[v], -cost, 0);
	head[v] = cnt++;
}
int q[MAXN*MAXN], vis[MAXN], dis[MAXN], fa[MAXN];
int pos[MAXN];
int db[MAXN], dc[MAXN], need[MAXN][5];
double dx[MAXN], dy[MAXN];
int spfa(int s, int t)
{
	int bg = 0, ed = 0;
	memset(vis, 0, sizeof vis);
	memset(fa, -1, sizeof fa);
	memset(dis, 0x3f, sizeof dis);
	dis[s] = 0;
	q[ed++] = s;
	while (bg < ed)
	{
		int u = q[bg++];
		vis[u] = 0;
		for (int i = head[u]; ~i; i=ee[i].next)
		{
			int v = ee[i].to;
			if (ee[i].val > 0 && dis[v]>dis[u]+ee[i].cost)
			{
				dis[v]=dis[u]+ee[i].cost;
				fa[v] = u;
				pos[v] = i;
				if (!vis[v])
				{
					vis[v] = 1;
					q[ed++] = v;
				}
			}
		}
	}
	return fa[t] != -1;
}
int minCost_maxFlow()
{
	int flow = 0;
	int res = 0;
	int end = 3*n+1, stt = 0;
	while (spfa(stt, end))
	{
		int mn = 0x3fffffff;
		for (int i = end; i!= stt; i = fa[i])
		{
			if (mn > ee[pos[i]].val)
				mn = ee[pos[i]].val;
		}
		for (int i = end; i!=stt; i=fa[i])
		{
			ee[pos[i]].val -= mn;
			ee[pos[i]^1].val += mn;
		}
		flow += mn;
		res += dis[end]*mn;
	}
	return res;
}
void init()
{
	cnt = 0;
	memset(head, -1, sizeof head);
}
double distce(double x, double y)
{
	return sqrt(x*x + y*y);
}
double lenth[MAXN/3][MAXN/3];
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif
	int t;
	scanf("%d", &t);
	while(t--)
	{
		int res = 0;
		scanf("%d%d", &n, &m);
		scanf("%lf%lf", dx, dy);
		for (int i = 1; i< n; ++i)
		{
			scanf("%lf%lf%d%d", dx+i, dy+i, db+i, dc+i);
			for (int j = 0; j< m; ++j) scanf("%d", &need[i][j]);
		}
		for (int i = 1; i< n; ++i)
		for (int j = i+1; j< n; ++j)
		lenth[i][j] = lenth[j][i] = distce(dx[i]-dx[j], dy[i]-dy[j]);	
		for (int i = 0; i< m; ++i)
		{
			init();
			for (int j = 1; j<n; ++j)
			{
				add(0, j, 1, need[j][i]);
				add(j, j+n, 0, need[j][i]);
				add(j+n, 3*n+1, 0, need[j][i]);
				add(0, j+2*n, 0, need[j][i]);
				for (int k = 1; k< n; ++k)
				if (db[j]+dc[j]+lenth[j][k]<=db[k])
				add(j+2*n, k+n, 0, need[j][i]);
			}
			res += minCost_maxFlow();
		}	
		printf("%d\n", res);
	}
	return 0;
}

hdu 4067 Random Maze

能不能看出门道来建立数学模型才是这类问题的关键

推荐这篇博客 http://blog.sina.com.cn/s/blog_64675f540100tnwn.html 

hdu 3947 River Problem

类似 https://www.byvoid.com/blog/noi-2008-employee/

很有意思的题目，感觉上有些像差分约束。关键是为了保证流入流出的平衡，选择用每条边减去其子节点的边

#include<cstdio>
#include<algorithm>
#include<vector>
#include<cmath>
#include<set>
#pragma comment(linker, "/STACK:1024000000,1024000000")
//acm.hdu.edu.cn/showproblem.php?pid=3947
using namespace std;
const int MAXN = 200, inf = 0x3f3f3f3f;
int n, m;
struct _node
{
	int to, next, cost, val;
	_node(int a, int b, int c, int d): to(a),next(b),cost(c),val(d) {}
	_node() {}
}ee[MAXN*MAXN];
int head[MAXN], cnt, ppre[MAXN], wt[MAXN];
void add(int u, int v, int cost, int val = inf)
{
	ee[cnt] = _node(v, head[u], cost, val);
	head[u] = cnt++;
	ee[cnt] = _node(u, head[v], -cost, 0);
	head[v] = cnt++;
}
int q[MAXN*MAXN], vis[MAXN], dis[MAXN], fa[MAXN], pos[MAXN];
int spfa(int s, int t)
{
	int bg = 0, ed = 0;
	memset(vis, 0, sizeof vis);
	memset(fa, -1, sizeof fa);
	memset(dis, 0x3f, sizeof dis);
	dis[s] = 0;
	q[ed++] = s;
	while (bg < ed)
	{
		int u = q[bg++];
		vis[u] = 0;
		for (int i = head[u]; ~i; i=ee[i].next)
		{
			int v = ee[i].to;
			if (ee[i].val > 0 && dis[v]>dis[u]+ee[i].cost)
			{
				dis[v]=dis[u]+ee[i].cost;
				fa[v] = u;
				pos[v] = i;
				if (!vis[v])
				{
					vis[v] = 1;
					q[ed++] = v;
				}
			}
		}
	}
	return fa[t] != -1;
}
int minCost_maxFlow(int s, int t, int all)
{
	int flow = 0;
	int res = 0;
	while (spfa(s, t))
	{
		int mn = inf;
		for (int i = t; i!= s; i = fa[i])
		{
			if (mn > ee[pos[i]].val)
				mn = ee[pos[i]].val;
		}
		for (int i = t; i!=s; i=fa[i])
		{
			ee[pos[i]].val -= mn;
			ee[pos[i]^1].val += mn;
		}
		flow += mn;
		res += dis[t]*mn;
	}
	if (flow != all) return -1;
	return res;
}
vector<int> sn[MAXN];
void minit()
{
	cnt = 0;
	memset(head, -1, sizeof head);
	for (int i = 0; i < n+5; ++i) sn[i].clear();
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif
	int T, cs = 0, s, t, all;
	scanf("%d", &T);
	while ( T--)
	{
		printf("Case #%d: ", ++cs);
		scanf("%d", &n);
		minit();
		s = 0; t = n+2;
		all = 0;
		// add node 1 & n+1
		for (int i = 1; i< n; ++i)
		{
			int u, v, w;
			scanf("%d%d%d", &u, &v, &w);
			wt[u] = w;
			ppre[u] = v;
			sn[v].push_back(u);
		}
		scanf("%d", &m);
		while (m--)
		{
			int u, v, l, c;
			scanf("%d%d%d%d", &u, &v, &l, &c);
			add(u, v, c, l);
		}
		for (int i = 1; i<= n; ++i)
		{
			int val = wt[i];
			for (int j = 0, k=sn[i].size(); j<k; ++j)
				val -= wt[sn[i][j]], add(i, sn[i][j], 0);
			if (val >= 0)
				add(s, i, 0, val), all += val;
			else
				add(i, t, 0, -val);			
		}
		printf("%d\n", minCost_maxFlow(s, t, all));
	}
	return 0;
}