HDU 1317 XYZZY

题目链接：https://vjudge.net/problem/HDU-1317

题目大意

　　有 N 个房间，每个房间都有一定能量（有正有负），其中某些房间单向可达，你现在开局自带 100 能量，想从 1 号房间走到 N 号房间，并且要保证中途剩余能量恒为正，问是否可能？

分析

　　我一开始不知道题目中“quits in frustration”是什么意思，后来我才知道，这句话的意思是不一定有路能到达 N。（说清楚一点会死么。。。）

　　本来就是个判断有没有正环的题目，现在还要判断正环上的点能否到达 N，不能的话这个正环就没什么卵用。

　　可以先用 Floyd 先求任意两点的可达性，然后用 SPFA 判断有无能到达 N 的正环，没有再判断 N 上的值是否为正。

　　PS：此题数据并不强，我的第一版代码 AC 了，但后来我自己弄了个数据，本地 WA 了，最终搞了第二版代码，本地和 OJ 都 AC 了。

代码如下

#include <bits/stdc++.h>
using namespace std;
 
#define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define Rep(i,n) for (int i = 0; i < (int)(n); ++i)
#define For(i,s,t) for (int i = (int)(s); i <= (int)(t); ++i)
#define rFor(i,t,s) for (int i = (int)(t); i >= (int)(s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end())
#define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c 
#define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower);
#define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper);
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,0x3f,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}

template<typename T>
ostream &operator<<(ostream &out, vector<T> &v) {
    Rep(i, v.size()) out << v[i] << " \n"[i == v.size()];
    return out;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "\n";
    return out;
}

inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
    
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}

template<class T>
inline string toString(T x) {
    ostringstream sout;
    sout << x;
    return sout.str();
}

inline int toInt(string s) {
    int v;
    istringstream sin(s);
    sin >> v;
    return v;
}

//min <= aim <= max
template<typename T>
inline bool BETWEEN(const T aim, const T min, const T max) {
    return min <= aim && aim <= max;
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef vector< int > VI;
typedef vector< bool > VB;
typedef vector< char > VC;
typedef vector< double > VD;
typedef vector< string > VS;
typedef vector< LL > VL;
typedef vector< VI > VVI;
typedef vector< VB > VVB;
typedef vector< VS > VVS;
typedef vector< VL > VVL;
typedef vector< VVI > VVVI;
typedef vector< VVL > VVVL;
typedef pair< int, int > PII;
typedef pair< LL, LL > PLL;
typedef pair< int, string > PIS;
typedef pair< string, int > PSI;
typedef pair< string, string > PSS;
typedef pair< double, double > PDD;
typedef vector< PII > VPII;
typedef vector< PLL > VPLL;
typedef vector< VPII > VVPII;
typedef vector< VPLL > VVPLL;
typedef vector< VS > VVS;
typedef map< int, int > MII;
typedef unordered_map< int, int > uMII;
typedef map< LL, LL > MLL;
typedef map< string, int > MSI;
typedef map< int, string > MIS;
typedef set< int > SI;
typedef stack< int > SKI;
typedef queue< int > QI;
typedef priority_queue< int > PQIMax;
typedef priority_queue< int, VI, greater< int > > PQIMin;
const double EPS = 1e-8;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 1e9 + 7;
const int maxN = 1e2 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

struct Edge{
    int from, to;
};

istream& operator>> (istream& in, Edge &x) {
    in >> x.from >> x.to;
    return in;
}

struct Vertex{
    int value, w;
    VI next;
    
    void clear() {
        value = 0;
        next.clear();
    }
};

int T, N, M;
Vertex V[maxN];
vector< Edge > E;
bool reach[maxN][maxN];

void addEdge(Edge &x) {
    V[x.from].next.PB(E.size());
    E.PB(x);
}

void init() {
    For(i, 1, N) V[i].clear();
    E.clear();
    ms0(reach);
}

bool vis[maxN]; // 标记是否在队列里 
int cnt[maxN]; // 记录一个顶点的入队次数 
bool SPFA(int S) {
    QI Q;
    
    ms0(vis);
    ms0(cnt);
    Q.push(S);
    vis[S] = 1;
    ++cnt[S];
    
    while(!Q.empty()) {
        int tmp = Q.front(); Q.pop();
        vis[tmp] = 0;
        
        Rep(i, V[tmp].next.size()) {
            Edge &e = E[V[tmp].next[i]];
            if(!reach[e.to][N]) continue; // 只松弛能到达 N 的点 
            
            if(V[e.to].value < V[e.from].value + V[e.to].w) { // 松弛 
                V[e.to].value = V[e.from].value + V[e.to].w; 
                if(!vis[e.to]) { // 一旦一个节点被更新，它就要入队，以保证它的后继节点被更新 
                    if(++cnt[e.to] >= N) return true; // 一旦一个节点被更新 N 次以上，就说明有正环并且可以达到  
                    Q.push(e.to);
                    vis[e.to] = 1; 
                }
            }
        }
    }
    return V[N].value > 0;
}

void Floyd() {
    For(k, 1, N) {
        reach[k][k] = true;
        For(i, 1, N) {
            if(!reach[i][k]) continue;
            For(j, 1, N) {
                if(!reach[k][j]) continue;
                reach[i][j] = true;
            }
        }
    }
}

int main(){
    //freopen("MyOutput.txt","w",stdout);
    //freopen("input.txt","r",stdin);
    INIT();
    while(cin >> N && N != -1) {
        init();
        For(i, 1, N) {
            int m;
            Edge t;
            t.from = i;
            
            cin >> V[i].w >> m;
            For(i, 1, m) {
                cin >> t.to;
                addEdge(t);
                reach[t.from][t.to] = true;
            }
        }
        
        Floyd();
        
        V[1].value = 100;
        if(reach[1][N] && SPFA(1)) cout << "winnable\n";
        else cout << "hopeless\n";
    }
    return 0;
}

/*
7
0 1 2
-80 2 3 6
-10 1 4
20 1 5
10 1 3
-100 1 7
0 0
hopeless

7
0 1 2
-80 2 3 6
-10 1 4
20 1 5
10 1 3
-10 1 7
0 0
winnable

9
0 1 2
-80 2 3 6
-10 1 4
20 1 5
10 1 3
-10 2 7 9
10 1 8
10 1 6
0 0
winnable
*/