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;
}