上下界网络流专题

BZOJ 2324ZJOI2011营救皮卡丘

XJTU校赛 贪吃蛇

BZOJ 2324([ZJOI2011]营救皮卡丘)

给定n点m边无向图，用k个人从起点出发，一个人走一条路代价为路的长度Li，你希望按照0,1,2,…,n的顺序依次经过这些点，其中经过的定义是任何1人经过该点，问k个人最小的道路总和。

N ≤ 150, M ≤ 20 000, 1 ≤ K ≤ 10, Li ≤ 10 000

考虑每次只有1个人走1步，已经过t点，

则每次其中一人走向t+1点，距离是所在点到t+1点的最短路（不经过t+1以上的点）

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (300+10)
#define MAXM ((20000)*12+10)
#define eps (1e-3)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
class Cost_Flow
{
public:
int n,s,t;
int q[MAXM];
int edge[MAXM],next[MAXM],pre[MAXN],weight[MAXM],size;
int cost[MAXM];
void addedge(int u,int v,int w,int c)
{
edge[++size]=v;
weight[size]=w;
cost[size]=c;
next[size]=pre[u];
pre[u]=size;
//        cout<<u<<' '<<v<<' '<<w<<endl;
}
void addedge2(int u,int v,int w,int c){addedge(u,v,w,c),addedge(v,u,0,-c);}
bool b[MAXN];
int d[MAXN];
int pr[MAXN],ed[MAXN];
bool SPFA(int s,int t)
{
For(i,n) d[i]=INF,b[i]=0;
d[q[1]=s]=0;b[s]=1;
int head=1,tail=1;
while (head<=tail)
{
int now=q[head++];
Forp(now)
{
int &v=edge[p];
if (weight[p]&&d[now]+cost[p]<d[v])
{
d[v]=d[now]+cost[p];
if (!b[v]) b[v]=1,q[++tail]=v;
pr[v]=now,ed[v]=p;
}
}
b[now]=0;
}
return d[t]!=INF;
}
int totcost;

int CostFlow(int s,int t)
{
int maxflow=0;
while (SPFA(s,t))
{
int flow=INF;
for(int x=t;x^s;x=pr[x]) flow=min(flow,weight[ed[x]]);
totcost+=flow*d[t];
maxflow+=flow;
for(int x=t;x^s;x=pr[x]) weight[ed[x]]-=flow,weight[ed[x]^1]+=flow;
}
//        cout<<maxflow<<endl;
return totcost;
}
void mem(int n,int t)
{
(*this).n=n;
size=1;
totcost=0;
MEM(pre) MEM(next)
}
}S1;
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
int n,m,k;
ll f[MAXN][MAXN]={0};
int main()
{
//  freopen("bzoj2324.in","r",stdin);
//  freopen(".out","w",stdout);
n=read();m=read();k=read();
int s=2*(n+1)+1,t=s+1,S=t+1,T=S+1;
S1.mem(T,T);
const int inf = INF;

Rep(i,n+1) Rep(j,n+1) f[i][j]=INF;
Rep(i,n+1) f[i][i]=0;
For(i,m) {
int a=read(),b=read();
f[a][b]=f[b][a]=min(f[a][b],(ll)read());
}
Rep(k,n+1) Rep(i,n+1) Rep(j,n+1) {
if (k>=max(i,j)) continue;
f[i][j]=min(f[i][j],f[i][k]+f[k][j]);
}
//  Rep(i,n+1) {
//      Rep(j,n+1) cout<<f[i][j]<<' ';
//      cout<<endl;
//  }
For(i,n+1) {
S1.addedge2(S,i+n+1,1,0);
S1.addedge2(i,T,1,0);
S1.addedge2(i,i+n+1,inf,0);

}
Rep(i,n+1) {
Fork(j,i+1,n) {
S1.addedge2(i+1+n+1,j+1,1,f[i][j]);
}
}
S1.addedge2(s,1,k,0);
Fork(i,n+2,2*n+2) S1.addedge2(i,t,inf,0);
S1.addedge2(t,s,k,0);

cout<<S1.CostFlow(S,T)<<endl;

return 0;
}

XJTU校赛 贪吃蛇

将矩阵黑白染色，相邻点连边，由于贪吃蛇长度至少为3，容易用上下界网络流表示环和头尾都在边界的情况

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case %d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a
<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
class Cost_Flow
{
#define MAXN (300+10)
#define MAXM (20000+10)
public:
int n,s,t;
int q[MAXM];
int edge[MAXM],next[MAXM],pre[MAXN],weight[MAXM],size;
int cost[MAXM];
void addedge(int u,int v,int w,int c)
{
edge[++size]=v;
weight[size]=w;
cost[size]=c;
next[size]=pre[u];
pre[u]=size;
}
void addedge2(int u,int v,int w,int c){addedge(u,v,w,c),addedge(v,u,0,-c);}
bool b[MAXN];
int d[MAXN];
int pr[MAXN],ed[MAXN];
bool SPFA(int s,int t)
{
For(i,n) d[i]=INF,b[i]=0;
d[q[1]=s]=0;b[s]=1;
int head=1,tail=1;
while (head<=tail)
{
int now=q[head++];
Forp(now)
{
int &v=edge[p];
if (weight[p]&&d[now]+cost[p]<d[v])
{
d[v]=d[now]+cost[p];
if (!b[v]) b[v]=1,q[++tail]=v;
pr[v]=now,ed[v]=p;
}
}
b[now]=0;
}
return d[t]!=INF;
}
int totcost,maxflow;
int CostFlow(int s,int t)
{

while (SPFA(s,t))
{
int flow=INF;
for(int x=t;x^s;x=pr[x]) flow=min(flow,weight[ed[x]]);
totcost+=flow*d[t];
maxflow+=flow;
for(int x=t;x^s;x=pr[x]) weight[ed[x]]-=flow,weight[ed[x]^1]+=flow;
}
return totcost;
}
void mem(int n,int t)
{
(*this).n=n;
size=1;  maxflow=0;
totcost=0;
MEM(pre) MEM(next)
}
}S1;
int n,m;
char s[20][20];
int main()
{
//  freopen("bzoj4213.in","r",stdin);
//  freopen(".out","w",stdout);

n=0;
while(scanf("%s",s[n+1]+1)!=EOF) s[++n][0]=0;
m=strlen(s[1]+1);

int S=n*m+1,T=S+1;
int SS=T+1,TT=SS+1;
S1.mem(TT,TT);
int p=0;
For(i,n) For(j,m) {
if (s[i][j]=='#') continue;
++p;
int now = (i-1)*m+j;
if ((i+j)&1) {
S1.addedge2(SS,now,2,0);
S1.addedge2(S,TT,2,0);

if (i<n) S1.addedge2(now,now+m,1,0);
if (i>1) S1.addedge2(now,now-m,1,0);
if (j<m) S1.addedge2(now,now+1,1,0);
if (j>1) S1.addedge2(now,now-1,1,0);

if (i==1||i==n||j==1||j==m) S1.addedge2(now,T,1,1);

} else {
S1.addedge2(SS,T,2,0);
S1.addedge2(now,TT,2,0);
if (i==1||i==n||j==1||j==m) S1.addedge2(S,now,1,1);
}

}
S1.CostFlow(SS,TT);
if (S1.maxflow<p*2) {
S1.addedge2(T,S,INF,0);
S1.CostFlow(SS,TT);
}
if (S1.maxflow<p*2) puts("-1");
else cout<<S1.totcost/2<<endl;

return 0;
}