航空路线问题 【网络流24题】

输入输出样例

输入 #1复制

8 9
Vancouver
Yellowknife
Edmonton
Calgary
Winnipeg
Toronto
Montreal
Halifax
Vancouver Edmonton
Vancouver Calgary
Calgary Winnipeg
Winnipeg Toronto
Toronto Halifax
Montreal Halifax
Edmonton Montreal
Edmonton Yellowknife
Edmonton Calgary

输出 #1复制

7
Vancouver
Edmonton
Montreal
Halifax
Toronto
Winnipeg
Calgary

 

 思路

　　路径问题，首先拆点

　　标准路径问题网络流解法

　　只需要在S和1、2n和T之间连流量为2的边即可

　　ans就是 mincost - 2

　　最后用dfs枚举悔边输出路径

CODE

#include <bits/stdc++.h>
#define dbg(x) cout << #x << "=" << x << endl
#define eps 1e-8
#define pi acos(-1.0)

using namespace std;
typedef long long LL;
const int maxn = 1e5 +7;
const int inf  = 0x3f3f3f3f;

int n, m, s, t;
int head[maxn],pre[maxn],inq[maxn],dis[maxn];
int a[maxn];
int cnt = 1;
int path[2][maxn];
int mincost = 0, maxflow = 0;
struct edge {
    int u,to,nxt,w,c;
}e[maxn << 1];
int tot[5];

template<class T>inline void read(T &res)
{
    char c;T flag=1;
    while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
    while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
}

inline void BuildGraph(int u, int v, int w, int cost)
{
    e[++cnt] = (edge){u, v, head[u], w,  cost}, head[u] = cnt;
    e[++cnt] = (edge){v, u, head[v], 0, -cost}, head[v] = cnt;///反向边
}

queue < int > q;

bool SPFA(int x)
{
    memset(inq, 0, sizeof(inq));
    for(int i = s; i <= t; i++) {
        dis[i] = inf;
    }
    q.push(x);
    dis[x] = 0;
    inq[x] = 1;
    while(!q.empty()) {
        int u = q.front();
        q.pop();
        inq[u] = 0;
        for(int i = head[u]; i; i = e[i].nxt) {
            int v = e[i].to, w = e[i].c;
            if(e[i].w > 0) {
                if(dis[u] + w < dis[v]) {
                    dis[v] = dis[u] + w;
                    pre[v] = i;
                    if(!inq[v]) {
                        q.push(v);
                        inq[v] = 1;
                    }
                }
            }
        }
    }
    if(dis[t] == inf)
        return 0;
    return 1;
}

void MCMF()
{
    while(SPFA(s)) {
        int temp = inf;
        for(int i = pre[t]; i; i = pre[e[i].u]) {
            temp = min(temp, e[i].w);
        }
        for(int i = pre[t]; i; i = pre[e[i].u]) {
            e[i].w   -= temp;
            e[i^1].w += temp;
            mincost += e[i].c * temp;
            //printf("ans:%d
",ans);
        }
        maxflow += temp;
    }
}

map<string, int> mp;
string ss[200];

void dfs(int x, int opt) {
    //cout << "x:" << x << endl;
    path[opt][++tot[opt]] = x;
    for ( int i = head[x]; i; i = e[i].nxt ) {
        int v = e[i].to;
        if(i & 1)
            continue;
        if(e[i ^ 1].w) {
            --e[i ^ 1].w;
            dfs(v, opt);
            return;
        }
    }
}


int main()
{
    //freopen("data.txt", "r", stdin);
    int vv;
    read(n); read(vv);
    s = 0, t = n << 1 | 1;
    for( int i = 1; i <= n; ++i ) {
        string _s;
        cin >> _s;
        mp[_s] = i;
        ss[i] = _s;
        BuildGraph(i, i + n, 1, -1);
    }
    BuildGraph(s, 1, 2, 0); BuildGraph(n << 1, t, 2, 0);
    BuildGraph(1, 1 + n, 1, -1); BuildGraph(n, n << 1, 1, -1);
    for( int i = 1; i <= vv; ++i ) {
        string a, b;
        cin >> a >> b;
        int u = mp[a], v = mp[b];
        if(u > v)
            swap(u, v);
        BuildGraph(u + n, v, inf, 0);
    }
    MCMF();
    bool flag = 0;
    // dbg(maxflow);
    // dbg(-mincost);
    if(!maxflow) {
        puts("No Solution!");
        return 0;
    }
    dfs(1, 0);
    dfs(1, 1);
    cout << -mincost - 2 << endl;
    for ( int i = 1; i <= tot[0]; ++i ) {
        if(path[0][i] >= 1 && path[0][i] <= n) {
            cout << ss[path[0][i]] << endl;
        }
    }
    //dbg(tot[1]);
    for ( int i = tot[1]; i >= 1; --i ) {
        if(path[1][i] >= 1 && path[1][i] <= n) {
            if(!flag) {
                flag = 1;
                continue;
            }
            cout << ss[path[1][i]] << endl;
        }
    }
    return 0;
}