BZOJ 1927: [Sdoi2010]星际竞速 [上下界费用流]

1927: [Sdoi2010]星际竞速

题意：一个带权DAG，每个点恰好经过一次，每个点有曲速移动到他的代价，求最小花费

不动脑子直接上上下界费用流过了...

s到点连边边权为曲速的代价，一个曲速移动等价于走到t再从s重新开始

搜了下题解发现全是普通费用流...


和最小路径覆盖很像，只是连到i+n有权值

***

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define fir first
#define sec second
typedef long long ll;
const int N=2005, M=1e5+5, INF=1e9;
inline ll read(){
char c=getchar();ll x=0,f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
return x*f;
}

int n, m, s, t, tot, extra
, x, val, warp;
struct edge{int v, c, f, w, ne, lower;}e[M];
int cnt=1, h
;
inline void ins(int u, int v, int c, int w=0, int lower=0) {
e[++cnt] = (edge){v, c, 0, w, h[u], lower}; h[u]=cnt;
e[++cnt] = (edge){u, 0, 0, -w, h[v], lower}; h[v]=cnt;
}
int q
, head, tail, d
, inq
;
pair<int, int> pre
;
inline void lop(int &x) {if(x==N) x=1;}
bool spfa(int s, int t) {
memset(inq, 0, sizeof(inq));
memset(d, 0x3f, sizeof(d));
head=tail=1;
q[tail++]=s; d[s]=0; inq[s]=1;
pre[t].fir = -1;
while(head!=tail) {
int u=q[head++]; inq[u]=0; lop(head);
for(int i=h[u];i;i=e[i].ne) {
int v=e[i].v;
if(e[i].c > e[i].f && d[v]>d[u]+e[i].w) {
d[v]=d[u]+e[i].w;
pre[v]=make_pair(u, i);
if(!inq[v]) q[tail++]=v, inq[v]=1, lop(tail);
}
}
}
return pre[t].fir != -1;
}
int ek(int s, int t) {
int flow=0, cost=0, x;
while(spfa(s, t)) {
int f=INF;
for(int i=t; i!=s; i=pre[i].fir) x=pre[i].sec, f=min(f, e[x].c - e[x].f);
flow+=f; cost+=d[t]*f;
for(int i=t; i!=s; i=pre[i].fir) x=pre[i].sec, e[x].f+=f, e[x^1].f-=f;
}
return cost;
}

int main() {
//freopen("in","r",stdin);
freopen("starrace.in","r",stdin);
freopen("starrace.out","w",stdout);
n=read(); m=read(); s=0; t=n+n+1;
for(int i=1; i<=n; i++)
warp=read(), ins(s, i, 1, warp), ins(i, i+n, 0, 0, 1), extra[i]--, extra[i+n]++, ins(i+n, t, 1);
for(int i=1; i<=m; i++) {
int u=read(), v=read(), val=read();
if(u>v) swap(u, v);
ins(u+n, v, 1, val);
}
int ss=t+1, tt=t+2, sum=0; tot=tt;
for(int i=s; i<=t; i++) {
if(extra[i]>0) ins(ss, i, extra[i]), sum+=extra[i];
if(extra[i]<0) ins(i, tt, -extra[i]);
}
ins(t, s, INF);
int ans=ek(ss, tt);
printf("%d",ans);
}