HDU 1317 XYZZY 最短路 Floyed SPFA

/*
 * 做題總結
 *      1 : 一開始以爲只要判正環，就沒寫Floyed，用了鏈式前向星存的圖，懶得改了=.=
 *      2 : 找最長路，核心代碼要改爲大於號，且題目要求大於0。dis數組初始化爲0，dis[1] = 100
 *      3 : 建圖過於複雜了，我是先把所有的讀入存起來再處理的。暫時沒想到好的做法。
 *      4 : Floyed判連通性很妙，短小精悍=.=
 */
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e2 + 10;
const int MAXM = 1e5 + 10;
const int INF = 0x3f3f3f3f;

struct input{
    int w, m;
    set<int> s;
}ds[MAXN];
int n;
struct E{
    int to, w, nxt;
}edge[MAXM];
int tot;
int head[MAXN];
queue<int> q;
int dis[MAXN];
bool inq[MAXN];
int inqtot[MAXN];
bool F[MAXN][MAXN];

inline void clear(){
    for(int i = 0; i < MAXN; ++i) ds[i].s.clear();
    tot = 0;
    memset(head, -1, sizeof head);
    while(!q.empty()) q.pop();
    memset(dis, 0, sizeof dis);
    dis[1] = 100;
    memset(inq, false, sizeof inq);
    memset(inqtot, 0, sizeof inqtot);
    memset(F, false, sizeof F);
}

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 void 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[id] + w > 0){
                dis[to] = dis[id] + w;
                if(!inq[to]){
                    inq[to] = true;
                    ++inqtot[to];
                    q.push(to);
                    if(inqtot[to] > n) {
                        if(F[to][n] == true) cout << "winnable" << endl;
                        else cout << "hopeless" << endl;
                        return ;
                    }
                }
            }
        }
    }
    if(dis[n] > 0) cout << "winnable" << endl;
    else cout << "hopeless" << endl;
}

inline void floyed(){
    for(int k = 1; k <= n; ++k)
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= n; ++j)
                F[i][j] = F[i][j] || (F[i][k] & F[k][j]);
}

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);

    while(cin >> n){
        if(n == -1) break;
        clear();
        for(int i = 1; i <= n; ++i){
            cin >> ds[i].w >> ds[i].m;
            for(int j = 1, x; j <= ds[i].m; ++j){
                cin >> x;
                ds[i].s.insert(x);  //重複的不要
            }
        }
        for(int i = 1; i <= n; ++i)
            for(auto it : ds[i].s){
                addedge(i, it, ds[it].w);
                F[i][it] = true;
            }
        floyed();
        SPFA();
    }
    return 0;
}