給定雙向邊，求Dalian到Shanghai再到Xian，途中不經過統一的城市，求最短距離
費用流
源點往Shanghai連邊，城市進行拆點處理（流量為1，確保只經過一次），然後城市之間流量為1，費用為對應距離，最後Dalian和Xian與匯點連邊，跑費用流。如果流量為2那麼費用就是最小距離了。

#include <bits/stdc++.h>
#include <map>
using namespace std;

const int MAXN = 40010;
const int MAXM = 100000;
const int INF = 0x3f3f3f3f
 
;
struct Edge {
    int to,next,cap,flow,cost;
}edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N, sum, m, n, sh;//節點總個數，節點編號從0~N-1
map<string, int> G;

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++;
}
void init(/*int n*/) {
    G.clear();
    tol = 0;
    memset(head,-1,sizeof(head));
    scanf("%d", &m);
    sum = 0
 
;
    string s1, s2;
    int len;
    n = 0;
    for (int i = 1; i <= m; i++) {
        cin >> s1;
        cin >> s2;
        cin >> len;
        sum += len;
        if (!G[s1]) G[s1] = ++n;
        if (!G[s2]) G[s2] = ++n;
        addedge(G[s1]+20000, G[s2], 1, len);
        addedge(G[s2]+20000, G[s1], 1, len);
    }
    sh = G["Shanghai"];
    for (int i = 1; i <= n; i++) {
        if (i == sh) addedge(i, i+20000, 2, 0);
        else addedge(i, i+20000, 1, 0);
    }
    addedge(0, sh, 2, 0);
    addedge(G["Dalian"], MAXN-1, 1, 0);
    addedge(G["Xian"], MAXN-1, 1, 0);
    N = n;
}
bool spfa(int s,int t) {
    queue<int>q;
    for(int i = 0;i < MAXN;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存的是最小費用
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;
}

int main() {
    freopen("input.txt","r",stdin);
    int T;
    scanf("%d", &T);
    while (T--) {
        init();
        if (G["Dalian"] && G["Shanghai"] && G["Xian"]) {
            int cost = 0;
            int flow = minCostMaxflow(0, MAXN-1, cost);
            if (flow < 2) {
                printf("-1\n");
            } else {
                printf("%d\n", cost);
            }
        } else {
            printf("-1\n");
        }
    }
}