題目鏈接
題意
一張航空圖,從西向東,1-n個城市,城市之間有若干條航線,航線可以往返,求從1到n再從n到1最多經過多少個不同的城市,1和n可以經過兩次,其他城市都只能經過一次。
思路
經過次數,顯然要想到拆點和流量控制,將每個城市點拆成入點和出點i和i+n,1與1+n之間的容量是2,費用是0,n與n+n之間的容量是2,費用是0,其他的入點與出點之間的容量是1,費用是0,這樣就限制了1和n可以經過兩次,其它點只能經過一次;因爲求的是經過的最多城市,用最大費用最大流,建立源點S,匯點T,S連1容量爲2,費用爲0,n+n連T容量爲2費用爲0,航線連通的兩個城市i,j,加邊i+n與j容量爲1,費用爲-1,然後跑費用流。最大流如果不是2,則不能從1到n往返。
輸出路徑就從1的出點開始遍歷,正向一次反向一次,第一次走過的第二次要跳過。
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
const int N = 5e3 + 7;
const int M = 1e4 + 7;
const int maxn = 5e3 + 7;
typedef long long ll;
int maxflow, mincost;
struct Edge {
int from, to, cap, flow, cost;
Edge(int u, int v, int ca, int f, int co):from(u), to(v), cap(ca), flow(f), cost(co){};
};
struct MCMF
{
int n, m, s, t;
vector<Edge> edges;
vector<int> G[N];
int inq[N], d[N], p[N], a[N];//是否在隊列 距離 上一條弧 可改進量
void init(int n) {
this->n = n;
for (int i = 0; i < n; i++) G[i].clear();
edges.clear();
}
void add(int from, int to, int cap, int cost) {
edges.push_back(Edge(from, to, cap, 0, cost));
edges.push_back(Edge(to, from, 0, 0, -cost));
int m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
bool SPFA(int s, int t, int &flow, int &cost) {
for (int i = 0; i < N; i++) d[i] = INF;
memset(inq, 0, sizeof(inq));
d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
queue<int> que;
que.push(s);
while (!que.empty()) {
int u = que.front();
que.pop();
inq[u]--;
for (int i = 0; i < G[u].size(); i++) {
Edge& e = edges[G[u][i]];
if(e.cap > e.flow && d[e.to] > d[u] + e.cost) {
d[e.to] = d[u] + e.cost;
p[e.to] = G[u][i];
a[e.to] = min(a[u], e.cap - e.flow);
if(!inq[e.to]) {
inq[e.to]++;
que.push(e.to);
}
}
}
}
if(d[t] == INF) return false;
flow += a[t];
cost += d[t] * a[t];
int u = t;
while (u != s) {
edges[p[u]].flow += a[t];
edges[p[u]^1].flow -= a[t];
u = edges[p[u]].from;
}
return true;
}
int MinMaxflow(int s, int t) {
int flow = 0, cost = 0;
while (SPFA(s, t, flow, cost));
maxflow = flow; mincost = cost;
return cost;
}
};
map<int, string> mp;
map<string, int> id;
MCMF solve;
int n, m, s, t, tot = 0;
void dfs(int x)
{
cout << mp[x - n] << endl;
if(x == n + n) return;
for (int i = 0; i < solve.G[x].size(); i++) {
Edge& e = solve.edges[solve.G[x][i]];
if(e.flow != e.cap || solve.G[x][i] & 1) continue;
dfs(e.to + n); return;
}
}
void dfs1(int x) {
if(x == n + n) return;
for (int i = 0; i < solve.G[x].size(); i++) {
Edge& e = solve.edges[solve.G[x][i]];
if(e.flow != e.cap || solve.G[x][i] & 1) continue;
dfs1(e.to + n); cout << mp[x - n] << endl;
}
}
int main()
{
cin >> n >> m;
s = 0, t = 2 * n + 1;
solve.add(s, 1, 2, 0);
solve.add(n + n, t, 2, 0);
for (int i = 1; i <= n; i++) {
string ss;
cin >> ss;
id[ss] = ++tot;
mp[tot] = ss;
}
for (int i = 2; i <= n - 1; i++) solve.add(i, i + n, 1, 0);
solve.add(1, 1 + n, 2, 0); solve.add(n, n + n, 2, 0);//1和n可以經過兩次
for (int i = 1; i <= m; i++) {
string s1, s2;
cin >> s1 >> s2;
int id1 = id[s1], id2 = id[s2];
if(id1 > id2) swap(id1, id2);
if(id1 == 1 && id2 == n) solve.add(1 + n, n, 2, -1);//特判如果是從1到n則可以走兩次
else solve.add(id1 + n, id2, 1, -1);
}
solve.MinMaxflow(s, t);
if(maxflow != 2) {
printf("No Solution!\n");
return 0;
}
printf("%d\n", -mincost);
dfs(n + 1);
bool flag = 0;
for (int i = 0; i < solve.G[1 + n].size(); i++) {
Edge& e = solve.edges[solve.G[n + 1][i]];
if(e.flow != e.cap || solve.G[n + 1][i] & 1) continue;//流量不爲0且是正向邊輸出
if(!flag) {
flag = 1;
continue;
}
dfs1(e.to + n); break;
}
cout << mp[1] << endl;
}