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描述
There is a funny car racing in a city with n junctions and m directed roads.
The funny part is: each road is open and closed periodically. Each road
is associate with two integers (a, b), that means the road will be open
for a seconds, then closed for b seconds, then open for a seconds... All
these start from the beginning of the race. You must enter a road when
it‘s open, and leave it before it‘s closed again.
Your goal is to drive from junction s and arrive at junction t as early
as possible. Note that you can wait at a junction even if all its
adjacent roads are closed.
输入
There will be at most 30 test cases. The first line of each case contains four integers n, m, s, t (1<=n<=300, 1<=m<=50,000, 1<=s,t<=n). Each of the next m lines contains five integers u, v, a, b, t (1<=u,v<=n, 1<=a,b,t<=105), that means there is a road starting from junction u ending with junction v. It‘s open for a seconds, then closed for b seconds (and so on). The time needed to pass this road, by your car, is t. No road connects the same junction, but a pair of junctions could be connected by more than one road.
输出
For each test case, print the shortest time, in seconds. It‘s always possible to arrive at t from s.
样例输入
样例输出
题目来源
题解:将时间看成最短路的模型,上一层的时间已经是最优的解了,所以下一层在更新的结点也会是最优的,如果到达的当前边时间为 T，那么就要判断是否要在当前城市等待了。。
#include <iostream> #include <cstdio> #include <cstring> using namespace std; const int INF = 999999999; int n,m; struct Edge{ int v,a,b,t,next; }edge[50005]; int tot; int head[305]; void addEdge(int u,int v,int a,int b,int t,int &k){ edge[k].v = v,edge[k].a = a,edge[k].b = b,edge[k].t = t,edge[k].next = head[u],head[u]=k++; } void init(){ memset(head,-1,sizeof(head)); tot = 0; } int low[305]; bool vis[305]; int dijsktra(int s,int t){ memset(vis,false,sizeof(vis)); for(int i=1;i<=n;i++){ low[i] = INF; } low[s] = 0; vis[s] = true; for(int i=1;i<n;i++){ int MIN = INF; for(int j=1;j<=n;j++){ if(low[j]<MIN&&!vis[j]){ MIN = low[j]; s = j; } } vis[s] = true; for(int k=head[s];k!=-1;k=edge[k].next){ int v = edge[k].v,a=edge[k].a,b = edge[k].b,tim = edge[k].t ; if(a>=tim){ int t0 = low[s]%(a+b); if(t0+tim<=a) low[v] = min(low[s]+tim,low[v]); else low[v] = min(low[s]+a+b-t0+tim,low[v]); } } } return low[t]; } int main() { int s,t,cas=1; while(scanf("%d%d%d%d",&n,&m,&s,&t)!=EOF){ init(); for(int i=1;i<=m;i++){ int u,v,a,b,t; scanf("%d%d%d%d%d",&u,&v,&a,&b,&t); addEdge(u,v,a,b,t,tot); } printf("Case %d: %d\n",cas++,dijsktra(s,t)); } return 0; }
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原文地址：http://www.cnblogs.com/liyinggang/p/5789712.html
