题目链接 P2469 [SDOI2010]星际竞速
这道题的上下界在于两个,其中一个显而易见,但是另一个却比较的隐藏了。
- 每个点都要求被经过一次且恰好一次
- 你需要至少通过瞬移这个技能一次,因为需要跳到其中的一个星球上去
于是,就可以根据这两个上下界来进行有上下界最小费用可行流了。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 16e2 + 7, maxM = 4e4 + 7;
int N, M, s, ps, t, head[maxN], cnt;
struct Eddge
{
int nex, u, v; ll flow, cost;
Eddge(int a=-1, int _u=0, int _v=0, ll b=0, ll c=0):nex(a), u(_u), v(_v), flow(b), cost(c) {}
}edge[maxM];
inline void addEddge(int u, int v, ll f, ll c)
{
edge[cnt] = Eddge(head[u], u, v, f, c);
head[u] = cnt++;
}
inline void _add(int u, int v, ll f, ll c) { addEddge(u, v, f, c); addEddge(v, u, 0, -c); }
struct MaxFlow_MinCost
{
int pre[maxN], S, T; ll Flow[maxN], dist[maxN];
queue<int> Q;
bool inque[maxN];
inline bool spfa()
{
for(int i=0; i<=T; i++) { pre[i] = -1; dist[i] = INF; inque[i] = false; }
while(!Q.empty()) Q.pop();
Q.push(S); dist[S] = 0; inque[S] = true; Flow[S] = INF;
while(!Q.empty())
{
int u = Q.front(); Q.pop(); inque[u] = false;
ll f, w;
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].v; f = edge[i].flow; w = edge[i].cost;
if(f && dist[v] > dist[u] + w)
{
dist[v] = dist[u] + w;
Flow[v] = min(Flow[u], f);
pre[v] = i;
if(!inque[v])
{
inque[v] = true;
Q.push(v);
}
}
}
}
return ~pre[T];
}
inline ll EK()
{
ll sum_Cost = 0;
while(spfa())
{
int now = T, las = pre[now];
while(now ^ S)
{
edge[las].flow -= Flow[T];
edge[las ^ 1].flow += Flow[T];
now = edge[las].u;
las = pre[now];
}
sum_Cost += dist[T] * Flow[T];
}
return sum_Cost;
}
} MF;
inline void init()
{
cnt = 0; s = 0; ps = N + N + 1; t = ps + 1; MF.S = t + 1; MF.T = MF.S + 1;
for(int i=0; i<=MF.T; i++) head[i] = -1;
_add(s, ps, N, 0);
_add(MF.S, ps, 1, 0);
_add(s, MF.T, 1, 0);
for(int i=1; i<=N; i++)
{
_add(MF.S, N + i, 1, 0);
_add(i, MF.T, 1, 0);
}
}
int main()
{
scanf("%d%d", &N, &M);
init();
for(int i=1, Wi; i<=N; i++)
{
scanf("%d", &Wi);
_add(ps, i, 1, Wi);
_add(N + i, t, 1, 0);
}
for(int i=1, u, v, w; i<=M; i++)
{
scanf("%d%d%d", &u, &v, &w);
if(u > v) swap(u, v);
_add(N + u, v, 1, w);
}
_add(t, s, INF, 0);
printf("%lld\n", MF.EK());
return 0;
}