2017 ACM-ICPC 乌鲁木齐网络赛 Our Journey of Dalain Ends 费用流

题意：

你要从西安出发先到上海，然后再到大连，每个城市只能经过一次，问最短路。

题解：

费用流。

找一条从西安到上海的路，再找一条从大连到上海的路，就可以了。

对于从x到y的一条最短路，费用流只要源点连x，汇点连y，流量为1即可。

所以源点连西安容量1， 连大连容量1， 汇点连上海容量2即可，因为每个点只能经过一次，需要拆点。

之所以不能西安到上海，再上海到大连，是因为有可能费用流直接西安到大连，然后上海到上海。

代码：

#include <bits/stdc++.h>
#ifdef LOCAL
#define debug(x) cout<<#x<<" = "<<(x)<<endl;
#else
#define debug(x) 1;
#endif

#define chmax(x,y) x=max(x,y)
#define chmin(x,y) x=min(x,y)
#define lson id<<1,l,mid
#define rson id<<1|1,mid+1,r
#define lowbit(x) x&-x
#define mp make_pair
#define pb push_back
#define fir first
#define sec second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const int MOD = 1e9 + 7;
const double PI = acos (-1.);
const double eps = 1e-10;
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int MAXN = 1e5 + 5;

const int MAXM = 100000;
struct Edge {
    int to, next, cap, flow, cost;
} edge[MAXM];
int head[MAXN], tol;
int pre[MAXN], dis[MAXN];
bool vis[MAXN];
int N;//节点总个数，节点编号从 0~N-1
void init (int n) {
    N = n;
    tol = 0;
    for (int i = 0; i <= n; i++) head[i] = -1;
}
void addedge (int u, int v, int cap, int cost) {
    edge[tol].to = v;
    edge[tol].cap = cap;
    edge[tol].cost = cost;
    edge[tol].flow = 0;
    edge[tol].next = head[u];
    head[u] = tol++;
    edge[tol].to = u;
    edge[tol].cap = 0;
    edge[tol].cost = -cost;
    edge[tol].flow = 0;
    edge[tol].next = head[v];
    head[v] = tol++;
}
bool spfa (int s, int t) {
    queue<int>q;
    for (int i = 0; i < N; i++) {
        dis[i] = INF;
        vis[i] = false;
        pre[i] = -1;
    }
    dis[s] = 0;
    vis[s] = true;
    q.push (s);
    while (!q.empty() ) {
        int u = q.front();
        q.pop();
        vis[u] = false;
        for (int i = head[u]; i != -1; i = edge[i].next) {
            int v = edge[i].to;
            if (edge[i].cap > edge[i].flow &&
                    dis[v] > dis[u] + edge[i].cost) {
                dis[v] = dis[u] + edge[i].cost;
                pre[v] = i;
                if (!vis[v]) {
                    vis[v] = true;
                    q.push (v);
                }
            }
        }
    }
    if (pre[t] == -1) return false;
    else return true;
}
//返回的是最大流， cost 存的是最小费用
int minCostMaxflow (int s, int t, int &cost) {
    int flow = 0;
    cost = 0;
    while (spfa (s, t) ) {
        int Min = INF;
        for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to]) {
            if (Min > edge[i].cap - edge[i].flow)
                Min = edge[i].cap - edge[i].flow;
        }
        for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to]) {
            edge[i].flow += Min;
            edge[i ^ 1].flow -= Min;
            cost += edge[i].cost * Min;
        }
        flow += Min;
    }
    return flow;
}


map<string, int> q;
int main() {
#ifdef LOCAL
    freopen ("input.txt", "r", stdin);
#endif
    int T;
    cin >> T;
    while (T--) {
        q.clear();
        int m;
        scanf ("%d", &m);
        init (m * 2 + 5);
        int tot = 0;
        for (int i = 1; i <= m; i++) {
            string s, t;
            int cost;
            cin >> s >> t;
            scanf ("%d", &cost);
            if (q.find (s) == q.end() ) {
                q[s] = tot;
                tot += 2;
            }
            if (q.find (t) == q.end() ) {
                q[t] = tot;
                tot += 2;
            }
            addedge (q[s] + 1, q[t], 1, cost);
            addedge (q[t] + 1, q[s], 1, cost);
        }
        int st = tot, en = tot + 1;
        for (int i = 0; i <= tot; i += 2)
            addedge (i, i + 1, 1, 0);
        if (q.find ("Shanghai") == q.end() || q.find ("Xian") == q.end() || q.find ("Dalian") == q.end() ) {
            puts ("-1");
            continue;
        }
        //addedge(q["Shanghai"], q["Shanghai"] ^ 1, 1, 0);
        addedge (st, q["Dalian"], 1, 0);
        addedge (st, q["Xian"], 1, 0);
        //addedge(q["Shanghai"] , en, 2, 0);
        int cost;
        int flow = minCostMaxflow (st, q["Shanghai"], cost);
        if (flow == 2) printf ("%d\n", cost);
        else puts ("-1");
    }
    return 0;
}