题意:M台机器要生产n个糖果，糖果i的生产区间在(si, ti)，花费是k(pi-si)，pi是实际开始生产的时间机器,j从初始化到生产糖果i所需的时间Cij，花费是Dij,任意机器从生产糖果i到生产糖果j，需花费时间Eij,花费Fij求生产完所有糖果所需的最小时间？
题解:先把糖果拆点,s向i糖果连流量为1,费用为0,i+n糖果向t连流量为1,费用为0,n2+i机器向t连流量为1,费用为0的边,如果i能流向j,那么j连i+n一条边流量为1,费用为(ti-sj)k+dij,如果机器j能流向i+2n,那么连一条边流量为1,费用为(p-s)*k+d,假设i流向了j,那么说明j在i之后生产,而且保证了每台机器只生产一种糖果
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
template<typename T>
inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>
inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}
using namespace std;
const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=300+10,maxn=400000+10,inf=0x3f3f3f3f;
struct edge{
int to,Next,c;
int cost;
}e[maxn];
int cnt,head[N];
int s,t;
int dis[N],pre[N],path[N];
bool vis[N];
void add(int u,int v,int c,int cost)
{
e[cnt].to=v;
e[cnt].c=c;
e[cnt].cost=cost;
e[cnt].Next=head[u];
head[u]=cnt++;
e[cnt].to=u;
e[cnt].c=0;
e[cnt].cost=-cost;
e[cnt].Next=head[v];
head[v]=cnt++;
}
bool spfa()
{
memset(pre,-1,sizeof pre);
memset(dis,inf,sizeof dis);
memset(vis,0,sizeof vis);
dis[s]=0;
vis[s]=1;
queue<int>q;
q.push(s);
while(!q.empty())
{
int x=q.front();
q.pop();
vis[x]=0;
for(int i=head[x];~i;i=e[i].Next)
{
int te=e[i].to;
if(e[i].c>0&&dis[x]+e[i].cost<dis[te])
{
dis[te]=dis[x]+e[i].cost;
pre[te]=x;
path[te]=i;
if(!vis[te])q.push(te),vis[te]=1;
}
}
}
return pre[t]!=-1;
}
int n,m,k;
int mincostmaxflow()
{
int cost=0,flow=0;
while(spfa())
{
int f=inf;
for(int i=t;i!=s;i=pre[i])
if(e[path[i]].c<f)
f=e[path[i]].c;
flow+=f;
cost+=dis[t]*f;
for(int i=t;i!=s;i=pre[i])
{
e[path[i]].c-=f;
e[path[i]^1].c+=f;
}
}
if(flow!=n)return -1;
return cost;
}
void init()
{
memset(head,-1,sizeof head);
cnt=0;
}
int cans[N],cant[N],ti[N][N];
int main()
{
while(~scanf("%d%d%d",&n,&m,&k))
{
if(!n&&!m&&!k)break;
init();
s=2*n+m+1,t=2*n+m+2;
for(int i=1;i<=m;i++)add(2*n+i,t,1,0);
for(int i=1;i<=n;i++)
{
scanf("%d%d",&cans[i],&cant[i]);
add(s,i,1,0);add(i+n,t,1,0);
}
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
scanf("%d",&ti[i][j]);
for(int i=1;i<=n;i++)
{
for(int j=1,x;j<=m;j++)
{
scanf("%d",&x);
if(ti[i][j]<cant[i])
{
add(i,2*n+j,1,max(0,ti[i][j] - cans[i])*k+x);
}
}
}
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
scanf("%d",&ti[i][j]);
for(int i=1;i<=n;i++)
{
for(int j=1,x;j<=n;j++)
{
scanf("%d",&x);
if(cant[i] + ti[i][j] < cant[j] && i != j)add(j,n+i,1,max(0,cant[i] + ti[i][j] - cans[j])*k+x);
}
}
printf("%d\n",mincostmaxflow());
}
return 0;
}
/********************
3 2 1
4 7
2 4
8 9
4 4
3 3
3 3
2 8
12 3
14 6
-1 1 1
1 -1 1
1 1 -1
-1 5 5
5 -1 5
5 5 -1
********************/