UVA12661 有趣的赛车比赛

白书上的例题，n个点m条单向边，每条边周期性开放和关闭，时间分别为a，b
求s到t的最短路

首先对于a>cost的边，可以直接删掉
spfa，算dist的时候，加入等待的时间
然后，就没有然后了

写代码越来越模块化（chou）了

#include <queue>#include <cstdio>#include <vector>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int N = 100007;const int INF=0x3f3f3f3f;struct SPFA{    struct Edge{        int from,to,cost,a,b;        Edge(){}        Edge(int x,int y,int z,int a,int b):            from(x),to(y),cost(z),a(a),b(b){}        bool canPass(){return a>=cost;}//题目里的        void read(){scanf("%d%d%d%d%d",&from,&to,&a,&b,&cost);}    };    int n, m;    vector <Edge> edges;    vector <int > G[N];    inline void AddEdge(){        Edge e; e.read();        if (!e.canPass())return ;        edges.push_back(e);        int top = edges.size();        G[e.from].push_back(top-1);    }    inline void Init(int n,int m){        this -> n = n; this -> m = m;        edges.clear();        for (int i=0;i<=n;i++)G[i].clear();        for (int i=1;i<=m;i++) AddEdge();    }    bool inq[N];    int d[N],cnt[N];    inline int spfa(int s,int t){        queue<int> Q;        memset(inq, 0, sizeof(inq));        memset(cnt, 0, sizeof(cnt));        memset( d ,INF,sizeof( d ));        d[s] = 0;inq[s]=1;Q.push(s);        for (;!Q.empty();){            int u = Q.front();Q.pop();            inq[u]= 0;            for (int i=0;i<G[u].size();i++){                Edge &e = edges[G[u][i]];                int val = d[u],s = d[u] ;//                val %= (e.a + e.b);//特殊题目                if (val>e.a) s+=e.b-(val-e.a);//                else if (e.a<val+e.cost)s+=e.a+e.b-val;//                if (d[u]<INF&&s+e.cost<d[e.to]){                    d[e.to] = s + e.cost;                    if (!inq[e.to]){                        Q.push(e.to);inq[e.to] = 1;                        if (++cnt[e.to]>n)return INF;                    }                }            }        }        return d[t];    }}g;int main(){    //freopen("in.txt","r",stdin);    int n,m,s,t;    for (int T=0;~scanf("%d%d%d%d",&n,&m,&s,&t);){        g.Init(n,m,s,t);        int ans = g.spfa(s,t);        printf("Case %d: %d\n",++T,ans);    }    return 0;}

这波图论挂vjudge上了
http://vjudge.net/contest/107921#overview
密码cww