HDU 2923 Einbahnstrasse 最短路 SPFA

/*
 * 解題總結 :
 *      1 : 寫的有點複雜了，但整體還是蠻清晰的。
 *      2 : 還是一個map和一個變量分配id，建圖過程複雜了點。
 *      3 : 正着跑一遍最短路，然後反向建圖跑一遍最短路。
 */
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e3 + 10;
const int MAXM = 2e4 + 10;
const int INF = 0x3f3f3f3f;

struct input{
    int u, v, w;
    string midsts;
}ds[MAXM];
struct E{
    int to, w, nxt;
}edge[MAXM];
queue<int> Q;
map<string, int> M;
int tot, head[MAXN], dis[MAXN], inqtot[MAXN], e[MAXN], n, C, R, tim;
bool inq[MAXN];

inline void clear(){
    while(!Q.empty()) Q.pop();
    tot = 0;
    memset(head, -1, sizeof head);
    memset(dis, INF, sizeof dis);
    dis[1] = 0;
    memset(inqtot, 0, sizeof inqtot);
    memset(inq, false, sizeof inq);
}

inline void addedge(int u, int v, int w){
    edge[tot].to = v;
    edge[tot].w = w;
    edge[tot].nxt = head[u];
    head[u] = tot++;
}

inline int getw(string ss){
    int m = ss.size(), ans = 0;
    for(int i = 0; i < m; ++i)
        if(isdigit(ss[i])){
            while(i < m && isdigit(ss[i])) ans = ans * 10 + ss[i++] - '0';
            return ans;
        }
}

inline bool SPFA(){
    Q.push(1);
    inq[1] = true;
    ++inqtot[1];
    while(!Q.empty()){
        int id = Q.front();
        Q.pop();
        inq[id] = false;
        for(int i = head[id]; ~i; i = edge[i].nxt){
            int to = edge[i].to;
            int w = edge[i].w;
            if(dis[id] + w < dis[to]){
                dis[to] = dis[id] + w;
                if(!inq[to]){
                    inq[to] = true;
                    ++inqtot[to];
                    Q.push(to);
                    if(inqtot[to] > n) return false;
                }
            }
        }
    }
    return true;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    //cout << setiosflags(ios::fixed) << setprecision(1); //保留小數點後1位
    //cout << setprecision(1); //保留1位有效數字
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);

    int cas = 1;
    while(cin >> n >> C >> R){
        if(n == 0 && C == 0 && R == 0) break;
        M.clear();
        tim = 0;
        clear();
        for(int i = 1; i <= C + 1; ++i){
            string name;
            cin >> name;
            if(M[name] == 0) M[name] = ++tim;
            e[i] = M[name];
        }
        for(int i = 1; i <= R; ++i){
            string strs, midstr, stre;
            cin >> strs >> midstr >> stre;
            if(M[strs] == 0) M[strs] = ++tim;
            if(M[stre] == 0) M[stre] = ++tim;
            int u = M[strs];
            int v = M[stre];
            int w = getw(midstr);
            if(midstr[0] == '<') addedge(v, u, w);
            if(midstr[midstr.size() - 1] == '>') addedge(u, v, w);
            ds[i].u = u;
            ds[i].v = v;
            ds[i].w = w;
            ds[i].midsts = midstr;
        }
        bool ok = SPFA();
        if(!ok) continue;
        int ans = 0;
        for(int i = 2; i <= C + 1; ++i) ans += dis[e[i]];
        clear();
        for(int i = 1; i <= R; ++i){
            string midstr = ds[i].midsts;
            if(midstr[0] == '<') addedge(ds[i].u, ds[i].v, ds[i].w);
            if(midstr[midstr.size() - 1] == '>') addedge(ds[i].v, ds[i].u, ds[i].w);
        }
        ok = SPFA();
        if(!ok) continue;
        for(int i = 2; i <= C + 1; ++i) ans += dis[e[i]];
        cout << cas++ << ". " << ans << endl;
    }
    return 0;
}