标签：style blog http color os io for ar
先上题目：
12661 Funny Car Racing
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.
Input
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.
Output
For each test case, print the shortest time, in seconds. It’s always possible to arrive at t from s.
Sample Input
3 2 1 3
1 2 5 6 3
2 3 7 7 6
3 2 1 3
1 2 5 6 3
2 3 9 5 6
Sample Output
Case 1: 20
Case 2: 9
题意：给出一个有向图，每条边都有三个值，打开时间间隔，关闭时间间隔，通过它需要多少时间，路的开关是循环往复的。给你起点和终点，问你最少需要多少时间才可以从起点到达终点(保证一定存在这种路)。
这其实是求最短路，用DIJ求最短路，然后对于从某个点开始可以到达的下一个点，在分析是否需要更新的时候，需要判断当前时刻路开了没有，如果开了看看够不够时间过去，然后根据这些情况更新最小值就可以了。这里需要注意的东西就是这事有向图不是无向图。
上代码：
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #define MAX 50002 5 #define ll long long 6 #define INF (((ll)1)<<60) 7 using namespace std; 8 9 typedef struct edge{ 10 int to,next; 11 ll a,b,c,s; 12 }edge; 13 14 edge e[MAX<<1]; 15 int p[302],tot; 16 int n,m,st,ed; 17 ll dis[MAX]; 18 19 void dij(){ 20 bool f[302]={0}; 21 for(int i=1;i<=n;i++){ 22 dis[i]=INF; 23 } 24 dis[st]=0; 25 for(int i=0;i<n;i++){ 26 int u; 27 ll dd=INF; 28 for(int j=1;j<=n;j++){ 29 if(!f[j] && dd>dis[j]) { 30 dd=dis[j]; 31 u=j; 32 } 33 } 34 if(dd==INF) break; 35 f[u]=1; 36 for(int j=p[u];j!=-1;j=e[j].next){ 37 ll t=dis[u]; 38 ll r=t%e[j].s; 39 if(r>=e[j].a){ 40 t+=(e[j].s-r)+e[j].c; 41 }else{ 42 if(e[j].a-r>=e[j].c) t+=e[j].c; 43 else t+=e[j].s-r+e[j].c; 44 } 45 dis[e[j].to]=min(t,dis[e[j].to]); 46 } 47 } 48 } 49 50 inline void add(int u,int v,int a,int b,int c){ 51 e[tot].to=v; e[tot].next=p[u]; 52 e[tot].a=a; e[tot].b=b; e[tot].c=c; e[tot].s=a+b; 53 p[u]=tot++; 54 } 55 56 int main() 57 { 58 int t,u,v,a,b,c; 59 //freopen("data.txt","r",stdin); 60 t=1; 61 while(scanf("%d %d %d %d",&n,&m,&st,&ed)!=EOF){ 62 memset(p,-1,sizeof(p)); 63 tot=0; 64 for(int i=0;i<m;i++){ 65 scanf("%d %d %d %d %d",&u,&v,&a,&b,&c); 66 if(c<=a){ 67 add(u,v,a,b,c); 68 //add(v,u,a,b,c); 69 } 70 } 71 dij(); 72 printf("Case %d: %lld\n",t++,dis[ed]); 73 74 } 75 return 0; 76 }
UVa - 12661 - Funny Car Racing,布布扣,bubuko.com
UVa - 12661 - Funny Car Racing
标签：style blog http color os io for ar
原文地址：http://www.cnblogs.com/sineatos/p/3910939.html
