After going through the receipts from your car trip through Europe this summer, you realised that the gas prices varied between the cities you visited. Maybe you could have saved some money if you were a bit more clever about where you filled your fu…

题目 用优先队列优化普通的广搜就可以过了. #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #include<queue> struct pq { int x,y,val; friend bool operator < (pq a,pq b) { return a.val > b.val; } }; priority_queue <p…

Rescue Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 14   Accepted Submission(s) : 7 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Angel was caught by the MOLIGPY…

Ignatius and the Princess I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10541    Accepted Submission(s): 3205Special Judge Problem Description The Princess has been abducted by the BEelzebub…

题目大意: 在n个点 m条边的无向图中 需要运送X单位牛奶 每条边有隐患L和容量C 则这条边上花费时间为 L+X/C 求从点1到点n的最小花费 优先队列维护 L+X/C 最小 广搜到点n #include <bits/stdc++.h> using namespace std; #define LL long long #define INF 0x3f3f3f3f #define mem(i,j) memset(i,j,sizeof(i)) #define inc(i,l,r) for(int…

Description After going through the receipts from your car trip through Europe this summer, you realised that the gas prices varied between the cities you visited. Maybe you could have saved some money if you were a bit more clever about where you fi…

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1336 There is an interesting calculator. It has 3 rows of buttons. Row 1: button 0, 1, 2, 3, ..., 9. Pressing each button appends that digit to the end of the display. Row 2: button +0, +1, +2, +3, ...…

题意:       悟空要救唐僧,中途有最多就把钥匙,和最多五条蛇,要求就得唐僧并且拿到所有种类的钥匙(两个1只拿一个就行),拿钥匙i之前必须拿到钥匙i-1,打蛇多花费一秒,问救出唐僧并且拿到所有种类的钥匙的最小花费时间. 思路:       应该两种方法吧,感觉很多都是用4维的标记,然后广搜的,我用的是3维的标记,然后优先队列广搜的,题目没啥难的,关键是读懂题,敲代码的时候细心点就行了.mark[x][y][key] 表示在x,y点是要是状态是key是否走过. #include<stdio.h…

题意:从s到m的最短时间.(“o"不能走,‘#’走一个花两个单位时间,‘.'走一个花一个单位时间) 思路:广搜和优先队列. #include <stdio.h> #include <string.h> #include <iostream> #include <queue> #define MAX 30 using namespace std; struct pos { int x; int y; int step; }; bool operato…

题目链接: http://acm.nyist.net/JudgeOnline/problem.php?pid=523 #include<iostream> #include<cstdio> #include<queue> using namespace std; /* 用普通搜索TLE,已知起点和终点,可以考虑双向广搜或A*算法加速搜索 双向广搜,一个方向从出发点向终点搜索,一个方向从终点向出发点搜索,搜索到相同的结点时,即找到最短路径. */ ; ][] = { ,,…