UVa 12661

到达一个点后讨论是否需要等待即可,需要等待则加上等待的时间

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;

const int maxn = 50010;

int n, m, s, t;

int h[305], cnt = 0;
struct E{
	int to, a, b, ti, next;
}e[maxn << 1];
void add(int u, int v, int a, int b, int ti){
	e[++cnt].to = v;
	e[cnt].a = a;
	e[cnt].b = b;
	e[cnt].ti = ti;
	e[cnt].next = h[u];
	h[u] = cnt;
}

int d[maxn];
void dij(int s){
	priority_queue<pii, vector<pii>, greater<pii> > q;
	memset(d, 0x3f, sizeof(d));
	d[s] = 0;
	q.push(make_pair(d[s], s));
	
	while(!q.empty()){
		pii p = q.top(); q.pop();
		int u = p.second;
		
		if(p.first > d[u]) continue;
		for(int i = h[u] ; i != -1 ; i = e[i].next){
			int v = e[i].to;
			
			int no = d[u] % (e[i].a + e[i].b);
			
			if(no <= e[i].a && (e[i].a - no) >= e[i].ti){ // 能通过
				if(d[v] > d[u] + e[i].ti){
					d[v] = d[u] + e[i].ti;
					q.push(make_pair(d[v], v));
				}
			} else{ // 等待
				int wait = (e[i].a + e[i].b) - no;
				if(d[v] > d[u] + e[i].ti + wait){
					d[v] = d[u] + e[i].ti + wait;
					q.push(make_pair(d[v], v));
				}
			}
		}
	}
}

ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }

int main(){
	int kase = 0;
	while(scanf("%d%d%d%d", &n, &m, &s, &t) != EOF){
		memset(h, -1, sizeof(h));
		cnt = 0;
		
		int u, v, a, b, ti;
		for(int i = 1 ; i <= m ; ++i){
			scanf("%d%d%d%d%d", &u, &v, &a, &b, &ti);
			if(ti > a) continue;
			add(u, v, a, b, ti);
		}
		
		dij(s);
		
		printf("Case %d: ", ++kase);
		printf("%d
", d[t]);
	}
	return 0;
}
原文地址:https://www.cnblogs.com/tuchen/p/14992586.html