题目链接
题意：
给你N个港口，有N艘船，它们分别在不同的矿厂，现在呢，它们要返回港口了，并且每个港口只能停靠一只船，而且呢船只要到了港口就不可以再往外跑了。现在给出K条边，是矿厂与矿厂之间的无向路径，然后呢有P条边，是港口与矿厂之间的无向路径。
思路：
在这里，我们需要把“进了港口就不能再出来了”这个关键点用好，我们肯定可以确定的是船只能在矿厂和矿厂之间移动，所以我们不妨让起点变成港口，并且再也没有港口的边了，我们直接让它们在矿厂与矿厂之间跑，就可以使得它们确定与该港口之间的最短路了，然后跑KM算法的最小权完美匹配（负边）。即可。
#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 <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)
#define Min3(a, b, c) min(a, min(b, c))
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 205;
int N, M, K, P;
struct KM
{
int n, mp[maxN][maxN], link_x[maxN], link_y[maxN];
bool vis_x[maxN], vis_y[maxN];
int que[maxN], top, fail, pre[maxN];
ll hx[maxN], hy[maxN], slk[maxN];
bool check(int i)
{
vis_x[i] = true;
if(link_x[i])
{
que[fail++] = link_x[i];
vis_y[link_x[i]] = true;
return true;
}
while(i)
{
link_x[i] = pre[i];
swap(i, link_y[pre[i]]);
}
return false;
}
void bfs(int S)
{
for(int i=1; i<=n; i++)
{
slk[i] = INF;
vis_x[i] = vis_y[i] = false;
}
top = 0; fail = 1; que[0] = S;
vis_y[S] = true;
while(true)
{
ll d = 0;
while(top < fail)
{
for(int i=1, j = que[top++]; i<=n; i++)
{
if(!vis_x[i] && slk[i] >= (d = hx[i] + hy[j] - mp[i][j]))
{
pre[i] = j;
if(d) slk[i] = d;
else if(!check(i)) return;
}
}
}
d = INF;
for(int i=1; i<=n; i++)
{
if(!vis_x[i] && d > slk[i]) d = slk[i];
}
for(int i=1; i<=n; i++)
{
if(vis_x[i]) hx[i] += d;
else slk[i] -= d;
if(vis_y[i]) hy[i] -= d;
}
for(int i=1; i<=n; i++)
{
if(!vis_x[i] && !slk[i] && !check(i)) return;
}
}
}
void Init()
{
for(int i=1; i<=n; i++)
{
link_x[i] = link_y[i] = 0;
hy[i] = slk[i] = 0;
vis_y[i] = false;
}
for(int i=1; i<=n; i++)
{
hx[i] = 0;
for(int j=1; j<=n; j++)
{
if(hx[i] < mp[i][j]) hx[i] = mp[i][j];
}
}
}
void solve()
{
n = N;
Init();
for(int i=1; i<=n; i++) bfs(i);
ll ans = 0;
for(int i=1; i<=n; i++) ans += hx[i] + hy[i];
printf("%lld\n", -ans);
}
}km;
int id[maxN], _Index, head[maxN], cnt;
struct Eddge
{
int nex, to, val;
Eddge(int a=-1, int b=0, int c=0):nex(a), to(b), val(c) {}
}edge[maxN * maxN];
inline void addEddge(int u, int v, int w)
{
edge[cnt] = Eddge(head[u], v, w);
head[u] = cnt++;
}
inline void _add(int u, int v, int w) { addEddge(u, v, w); addEddge(v, u, w); }
vector<pair<int, int>> E[maxN];
int dis[maxN];
struct node
{
int id, val;
node(int a=0, int b=0):id(a), val(b) {}
friend bool operator < (node e1, node e2) { return e1.val > e2.val; }
};
priority_queue<node> Q;
inline void Dijkstra(int S)
{
while(!Q.empty()) Q.pop();
for(int i=1; i<=M; i++) dis[i] = INF;
int u, val;
for(pair<int, int> it : E[S])
{
u = it.first; val = it.second;
Q.push(node(u, val));
dis[u] = val;
}
node now;
while(!Q.empty())
{
now = Q.top(); Q.pop();
u = now.id;
if(dis[u] < now.val) continue;
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(dis[v] > dis[u] + edge[i].val)
{
dis[v] = dis[u] + edge[i].val;
Q.push(node(v, dis[v]));
}
}
}
for(int i=1; i<=N; i++) km.mp[S][i] = -dis[i];
}
inline void init()
{
_Index = cnt = 0;
for(int i=1; i<=M; i++)
{
id[i] = 0;
head[i] = -1;
}
for(int i=1; i<=N; i++) for(int j=1; j<=N; j++) km.mp[i][j] = -INF;
}
int main()
{
while(scanf("%d%d%d%d", &N, &M, &K, &P) != EOF)
{
init();
for(int i=1, ID; i<=N; i++)
{
scanf("%d", &ID);
id[ID] = ++_Index;
}
for(int i=1; i<=M; i++) if(!id[i]) id[i] = ++_Index;
for(int i=1, u, v, w; i<=K; i++)
{
scanf("%d%d%d", &u, &v, &w);
u = id[u]; v = id[v];
_add(u, v, w);
}
for(int i=1, u, v, w; i<=P; i++)
{
scanf("%d%d%d", &u, &v, &w);
v = id[v];
E[u].push_back(MP(v, w));
}
for(int i=1; i<=N; i++) Dijkstra(i);
km.solve();
}
return 0;
}
来源：CSDN
作者：Andres_Lionel
链接：https://blog.csdn.net/qq_41730082/article/details/103793517