HDU 2686 Matrix(最大費用最大流+拆點)
和POJ3422一樣
刪掉K把匯點與源點的容量改為2（因為有兩個方向的選擇）即可
#include
#include
#include
#include
#include
#include
const int maxn = 20000;
const int maxm = 800000;
const int inf = 1e8;
const int INF = 0x3f3f3f3f;
#define MIN INT_MIN
#define MAX 1e6
#define LL long long
#define init(a) memset(a,0,sizeof(a))
#define FOR(i,a,b) for(int i = a;ib)?(a):(b)
#define min(a,b) (a>b)?(b):(a)
using namespace std;
struct node
{
int u,v,w,cap,next;
} edge[maxm];
int pre[maxn],dis[maxn],head[maxn],cnt;
bool vis[maxn];
int n;
void add(int u,int v,int c,int cap)
{
edge[cnt].u=u;
edge[cnt].v=v;
edge[cnt].w=c;
edge[cnt].cap=cap;
edge[cnt].next=head[u];
head[u]=cnt++;
edge[cnt].u=v;
edge[cnt].v=u;
edge[cnt].w=-c;
edge[cnt].cap=0;
edge[cnt].next=head[v];
head[v]=cnt++;
}
int spfa(int s,int t)
{
queueq;
while(q.empty()==false) q.pop();
q.push(s);
memset(vis,0,sizeof(vis));
memset(pre,-1,sizeof(pre));
FOR(i,s,t+1)
dis[i] = -1;//求最長路dis數組初始化為-1
dis[s]=0;
vis[s] = 1;
while(!q.empty())
{
int u=q.front();
q.pop();
vis[u] = 0;
for(int i=head[u]; i!=-1; i=edge[i].next)
{
if(edge[i].cap && dis[edge[i].v] < (dis[u]+edge[i].w))//求最長路
{
dis[edge[i].v] = dis[u]+edge[i].w;
pre[edge[i].v] = i;
if(!vis[edge[i].v])
{
vis[edge[i].v]=1;
q.push(edge[i].v);
}
}
}
}
if(dis[t] != -1)//------------------忘改了。。
return 1;
else
return 0;
}
int MinCostMaxFlow(int s,int t)
{
int flow=0,cost=0;
while(spfa(s,t))
{
int df = inf;
for(int i = pre[t]; i!=-1; i=pre[edge[i].u])
{
if(edge[i].cap
