UVa 12661 Funny Car Racing (dijkstra)
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题意:给定一个有向图,每条路有5个整数修饰,u, v, a, b, t,表示起点为u,终点为v,打开时间a,关闭时间为b,通过时间为t,打开关闭是交替进行的,
问你从s到t最短时间是多少。
析:使用dijkstra算法,从每个结点出发,求最短路,并维护时间的最小值,这个可以用优先队列,然后考虑能不能通过这条路,如果t<a,可以在输入时处理。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #define debug() puts("++++") #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-5; const int maxn = 300 + 10; const int mod = 1e6; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } struct Node{ int to, a, b, t; }; vector<Node> G[maxn]; int d[maxn]; int dijkstra(int s, int ttt){ priority_queue<P, vector<P>, greater<P> > pq; pq.push(P(0, s)); fill(d, d+n+1, INF); d[s] = 0; while(!pq.empty()){ P p = pq.top(); pq.pop(); if(p.second == ttt) return p.first; int v = p.second; if(d[v] < p.first) continue; for(int i = 0; i < G[v].size(); ++i){ Node &u = G[v][i]; int t = p.first % (u.a+u.b); if(t + u.t <= u.a && d[u.to] > d[v] + u.t){ d[u.to] = d[v] + u.t; pq.push(P(d[u.to], u.to)); } else if(t + u.t > u.a){ int tt = u.a + u.b - t + u.t; if(d[u.to] > d[v] + tt){ d[u.to] = d[v] + tt; pq.push(P(d[u.to], u.to)); } } } } return 0; } int main(){ int kase = 0; int s, t; while(scanf("%d %d %d %d", &n, &m, &s, &t) == 4){ int u, v, a, b, tt; for(int i = 1; i <= n; ++i) G[i].clear(); while(m--){ scanf("%d %d %d %d %d", &u, &v, &a, &b, &tt); if(tt > a) continue; G[u].push_back((Node){v, a, b, tt}); } printf("Case %d: %d\n", ++kase, dijkstra(s, t)); } return 0; }
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