【spoj371】【最小费用最大流】BOXES

比较简单的一道费用流。

建模方法很直观:

若sum[i] > 1,连边(s,i,sum[i] - 1,0)

若sum[i] = 0,连边(i,t,1,0)

对于每个格子i的相邻格子j连边:(i,j,inf,1)

一次最小费用最大流即可。

代码:

#include<cstdio>
#include<cstring>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 1000 + 10;
const int maxm = 50000;
const int maxque = 200000;
struct Edge
{
	int pos,d,w,c;
	int next;
}E[maxm];
int head[maxn];
int sum[maxn],que[maxque];
int dis[maxn],pre[maxn];
bool vis[maxn];
int T,NE,n;
int s,t;
void init()
{
	freopen("spoj371.in","r",stdin);
	freopen("spoj371.out","w",stdout);
}

void insert(int u,int v,int c,int w)
{
	E[NE].c = c;E[NE].w = w;E[NE].pos = v;E[NE].d = u;
	E[NE].next = head[u];head[u] = NE++;
	E[NE].c = 0;E[NE].w = -w;E[NE].pos = u;E[NE].d = v;
	E[NE].next = head[v];head[v] = NE++;
}

bool spfa()
{
	memset(pre,-1,sizeof(pre));
	memset(vis,0,sizeof(vis));
	memset(dis,0x3f,sizeof(dis));
	dis[s] = 0;
	int l = 0,r = 0;
	que[r++] = s;
	vis[s] = true;
	while(l < r)
	{
		int u = que[l++];
		vis[u] = false;
		for(int i = head[u];i != -1;i = E[i].next)
		{
			int v = E[i].pos;
			if(E[i].c && dis[u] + E[i].w < dis[v])
			{
				dis[v] = dis[u] + E[i].w;
				pre[v] = i;
				if(!vis[v])
				{
					vis[v] = true;
					que[r++] = v;
				}
			}
		}
	}
	if(dis[t] == inf)return false;
	else return true;
}

int MCMF()
{
	int ret = 0,flow = 0;
	while(spfa())
	{
		int u = t;
		int min = inf;
		while(u != s)
		{
			if(E[pre[u]].c < min)min = E[pre[u]].c;
			u = E[pre[u]].d;
		}
		flow += min;
		u = t;
		while(u != s)
		{
			E[pre[u]].c -= min;
			E[pre[u]^1].c += min;
			u = E[pre[u]].d;
		}
		ret += dis[t];
	}
	return ret;
}

void solve()
{
	memset(E,0,sizeof(E));
	memset(head,-1,sizeof(head));
	s = 0,t = n + 1;
	NE = 0;
	for(int i = 1;i <= n;i++)
	{
		if(sum[i] > 1)insert(s,i,sum[i] - 1,0);
		if(sum[i] == 0)insert(i,t,1,0);
		if(i != 1 && i != n)
		{
			insert(i,i - 1,inf,1);
			insert(i,i + 1,inf,1);
		}
		else
		{
			if(i == 1)
			{
				insert(i,n,inf,1);
				insert(i,i + 1,inf,1);
			}
			else
			{
				insert(i,i - 1,inf,1);
				insert(i,1,inf,1);
			}
		}
	}
	printf("%d\n",MCMF());
}

void readdata()
{
	scanf("%d",&T);
	while(T--)
	{
		memset(sum,0,sizeof(sum));
		scanf("%d",&n);
		for(int i = 1;i <= n;i++)
		{
			scanf("%d",&sum[i]);
		}
		solve();
	}
}

int main()
{
	init();
	readdata();
	return 0;
}