2014在百度之星资格赛的第二个问题Disk Schedule

事实上，我认为它可以用来费用流问题。但光建地图上加班。

。。不科学啊。。

。

因副作用太大，否则，必然在。最后，想啊想，或者使用dp对。。。。

别想了一维dp。。。

。我不知道我是怎么想。无论如何，这是ac该

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
using namespace std;
#ifdef _WIN32
#define i64 __int64
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
#define i64 long long
#define out64 "%lld\n"
#define in64 "%lld"
#endif
/************ for topcoder by zz1215 *******************/
#define foreach(c,itr)  for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)
#define FF(i,a)         for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a,b)      for( int i = (a) ; i >= (b) ; i --)
#define S64(a)          scanf(in64,&a)
#define SS(a)           scanf("%d",&a)
#define LL(a)           ((a)<<1)
#define RR(a)           (((a)<<1)+1)
#define pb              push_back
#define pf              push_front
#define X               first
#define Y               second
#define CL(Q)           while(!Q.empty())Q.pop()
#define MM(name,what)   memset(name,what,sizeof(name))
#define MC(a,b)		memcpy(a,b,sizeof(b))
#define MAX(a,b)        ((a)>(b)?

(a):(b))
#define MIN(a,b)        ((a)<(b)?

(a):(b))
#define read            freopen("in.txt","r",stdin)
#define write           freope("out.txt","w",stdout)

const int inf = 0x3f3f3f3f;
const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-9;
const double pi = acos(-1.0);
const int maxn = 1111;

int dp[maxn];
int t[maxn];
int s[maxn];
int m,n;

int pas[maxn];

int pass(int a,int b)
{
if(a>=b) return 0;
return pas[b]-pas[a];
}

int find(int x,int y)
{
return min(abs(x-y),360-abs(x-y));
}

int dis(int a,int b)
{
return abs(t[b]-t[a])*400 + find(s[b],s[a]);
}

int main()
{
cin>>m;
for(int tt=1;tt<=m;tt++)
{
cin>>n;
t[0] = 0;
s[0] = 0;
for(int i=1;i<=n;i++)
{
SS(t[i]);
SS(s[i]);
}

pas[0] = 0;
for(int i=1;i<=n;i++)
{
pas[i] = pas[i-1]+dis(i-1,i);
}

dp[0] = 0;

for(int i=1;i<=n;i++)
{
dp[i] = dis(0,i) + pass(0,i-1);
for(int j=1;j<i;j++)
{
dp[i] = min(dp[i], dp[j] + dis(j,i) + dis(j-1,j+1)+pass(j+1,i-1) );
dp[i] = min(dp[i], dp[j] + dis(j,i)+ pass(j-1,i-1));
dp[i] = min(dp[i], dp[j] + dis(j-1,i) + pass(j,i-1));
}
}
dp
+=dis(n-1,n)+n*10;
cout<<dp
<<endl;
}
return 0;
}

接下来是费用流tle代码。大概意思就是把数据点的那条边的费用变成一个非常小的负数。那么，一定要会选择那条边，不然得到的一定不是最小费用流，这样就起了限制作用

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
using namespace std;
#ifdef _WIN32
#define i64 __int64
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
#define i64 long long
#define out64 "%lld\n"
#define in64 "%lld"
#endif
/************ for topcoder by zz1215 *******************/
#define foreach(c,itr)  for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)
#define FF(i,a)         for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a,b)      for( int i = (a) ; i >= (b) ; i --)
#define S64(a)          scanf(in64,&a)
#define SS(a)           scanf("%d",&a)
#define LL(a)           ((a)<<1)
#define RR(a)           (((a)<<1)+1)
#define pb              push_back
#define pf              push_front
#define X               first
#define Y               second
#define CL(Q)           while(!Q.empty())Q.pop()
#define MM(name,what)   memset(name,what,sizeof(name))
#define MC(a,b)        memcpy(a,b,sizeof(b))
#define MAX(a,b)        ((a)>(b)?(a):(b))
#define MIN(a,b)        ((a)<(b)?(a):(b))
#define read            freopen("in.txt","r",stdin)
#define write           freope("out.txt","w",stdout)

const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-9;
const double pi = acos(-1.0);
const int maxn = 2222;
const int add = 1111;

int m,n;
int t[add];
int s[add];

struct zz
{
int from;
int to;
i64 c;
i64 cost;
int id;
}zx;

int start = 0;
int end = 2221;
i64 cc = 100000000;

vector<zz>g[maxn];
i64 way[maxn];
int bid[maxn];
bool inq[maxn];

void link(int now,int to,int c,i64 cost,int bc=0)
{
zx.from = now;zx.to=to;zx.c=c;zx.cost=cost;
zx.id = g[zx.to].size();
g[zx.from].pb(zx);
swap(zx.from,zx.to);
zx.c=bc;zx.cost=-cost;
zx.id = g[zx.to].size()-1;
g[zx.from].pb(zx);
return ;
}

bool spfa()
{
for(int i=0;i<maxn;i++)
{
way[i]=inf64;   // 最小费用流
}
way[start]=0;
deque<int>q;
MM(inq,false);
inq[start]=true;
q.pb(start);
i64 now,to,temp;
while(!q.empty())
{
now = q.front();
q.pop_front();
for(int i=0;i<g[now].size();i++)
{
to = g[now][i].to;
if(g[now][i].c>0)
{
temp = g[now][i].cost+way[now];
if(temp<way[to])
{
bid[to] = g[now][i].id;
way[to] = temp;
if(!inq[to])
{
inq[to]=true;
if(q.empty() || way[to]<way[q.front()]) q.pf(to);   // 最小费用流
else
q.pb(to);
}
}
}
}
inq[now]=false;
}
return way[end]!=inf64;
}
pair<i64,i64> dfs(i64 flow = inf64,i64 to = end)
{
if(to == start) return make_pair(flow,0);
i64 now=g[to][bid[to]].to;
int id = g[to][bid[to]].id;
pair<i64,i64>temp=dfs(min(flow,g[now][id].c),now);
g[now][id].c-=temp.X;
g[to][bid[to]].c+=temp.X;
temp.Y+=temp.X*g[now][id].cost;
return temp;
}

i64 gao()
{
i64 ans=0;
spfa();
ans+=dfs().Y;
spfa();
ans+=dfs().Y;
return ans;
}

int find(int x,int y)
{
return min( abs(x-y), 360 - abs(x-y));
}

int dis(int at,int as,int bt,int bs)
{
return (abs(bt-at))*400 + find(as,bs);
}

void build()
{
for(int i=0;i<maxn;i++)
{
g[i].clear();
}
for(int i=1;i<=n;i++)
{
link(start,i,1,dis(0,0,t[i],s[i]));
}
link(start,end,1,dis(0,0,t
,s
));
for(int i=1;i<=n;i++)
{
link(i,i+add,1,-cc);
for(int j=i+1;j<=n;j++)
{
link(i+add,j,1,dis(t[i],s[i],t[j],s[j]));
}
}
for(int i=1;i<=n;i++)
{
link(i+add,end,1,dis(t[i],s[i],t
,s
));
}
return ;
}

int main()
{
cin>>m;
for(int i=1;i<=m;i++)
{
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>t[i]>>s[i];
}
build();
i64 ans = gao();
ans += cc*n;
ans += 10*n;
cout<<ans<<endl;
}
return 0;
}