Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 19122   Accepted: 7366 Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for brick…

 POJ 1789 -- Truck History Prim求分母的最小.即求最小生成树 #include<iostream> #include<cstring> #include<algorithm> using namespace std; + ; ; int n;//有几个卡车 ]; int d[maxn];//记录编号的数值 int Edge[maxn][maxn]; int dist[maxn]; void prim() { ; //加入源点 dist[]…

点击打开链接 Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 15235   Accepted: 5842 Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or fo…

(￣▽￣)" #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<vector> #include<queue> using namespace std; ; const int INF=10e8; int n,k,minn; ]; int c[MAXN][MAXN],lc[MAXN]; bool vis[MAX…

题目传送门 题意:给出n个长度为7的字符串,一个字符串到另一个的距离为不同的字符数,问所有连通的最小代价是多少 分析:Kuskal/Prim: 先用并查集做,简单好写,然而效率并不高,稠密图应该用Prim而且要用邻接矩阵,邻接表的效率也不高.裸题但题目有点坑爹:( Kruskal: #include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <…

题目连接 http://poj.org/problem?id=1789 Truck History Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bricks. The company has its own code describing each…

链接: http://poj.org/problem?id=1789 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22010#problem/G Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 14950   Accepted: 5714 Description Advanced Cargo Movement, Ltd. uses…

Truck History 题目链接: http://acm.hust.edu.cn/vjudge/contest/124434#problem/E Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bricks. The company has its…

Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 21518   Accepted: 8367 Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for brick…

题目链接:POJ 1789 Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bricks. The company has its own code describing each type of a truck. The code is simply…

题目链接:http://poj.org/problem?id=1789 Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bricks. The company has its own code describing each type of a truck. The code…

Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bricks. The company has its own code describing each type of a truck. The code is simply a string of ex…

模板题 题目:http://poj.org/problem?id=1789 题意:有n个型号,每个型号有7个字母代表其型号,每个型号之间的差异是他们字符串中对应字母不同的个数d[ta,tb]代表a,b之间的差异数问1/Σ(to,td)d(to,td)最大值 prime: #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack>…

题目链接:http://poj.org/problem?id=1789 大意: 不同字符串相同位置上不同字符的数目和是它们之间的差距.求衍生出全部字符串的最小差距. #include<stdio.h> #include<math.h> #include<algorithm> using namespace std; ; int cnt; int pre[MAXN]; ]; struct Edge { int from, to; int val; }edge[MAXN *…

题目链接:http://poj.org/problem?id=1789 题目意思:给出 N 行,每行7个字符你,统计所有的 行 与 行 之间的差值(就是相同位置下字母不相同),一个位置不相同就为1,依次累加.问最终的差值最少是多少. 额.....题意我是没看懂啦= =......看懂之后,就转化为最小生成树来做了.这是一个完全图,即每条边与除它之外的所有边都连通.边与边的权值是通过这个差值来算出来的. #include <iostream> #include <cstdio> us…

主题链接:http://poj.org/problem?id=1789 思维:一个一个点,每两行之间不懂得字符个数就看做是权值.然后用kruskal算法计算出最小生成树 我写了两个代码一个是用优先队列写的.可是超时啦,不知道为什么.希望有人能够解答.后面用的数组sort排序然后才AC. code: 数组sort排序AC代码: #include<cstdio> #include<queue> #include<algorithm> #include<iostream…

kruskal思想,排序后暴力枚举从任意边开始能够组成的最小生成树 #include <cstdio> #include <algorithm> using namespace std; const int maxn = 101; const int maxe = maxn * maxn / 2; struct edge{ int f,t,c; bool operator <(edge e2)const { return c<e2.c; } }e[maxe]; int…

最小生成树,主要是题目比较难懂. #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; +; *+; int Father[Maxn]; struct Edge { int from,to,w; }edge[maxn]; int n,tot; ]; int Find(int x) { if(x!=Father[x]) Fa…

id=1789">Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 17610   Accepted: 6786 Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture,…

Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bricks. The company has its own code describing each type of a truck. The code is simply a string of ex…

Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 20768   Accepted: 8045 Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for brick…

思路: 一开始用Kruskal超时了,因为这是一个稠密图,边的数量最惨可能N^2,改用Prim. Prim是这样的,先选一个点(这里选1)作为集合A的起始元素,然后其他点为集合B的元素,我们要做的就是每次找到B中的一个点,满足这个点到A的权值是B到A的权值中最小的,然后我们把这个点加入到A,再更新B中的点到A的最小距离. 代码: #include<cstdio> #include<cmath> #include<cstring> #include<iostream…

http://poj.org/problem?id=1789 Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bricks. The company has its own code describing each type of a truck. Th…

http://poj.org/problem?id=1789 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 27597   Accepted: 10731 Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furni…

http://poj.org/problem?id=3126 搜索的时候注意 1:首位不能有0 2:可以暂时有没有出现在目标数中的数字 #include <cstdio> #include <cstring> #include <queue> using namespace std; const int maxn=1e4+5; const int inf=0x7fffffff; bool prim[maxn]; void judge(){ prim[2]=true; f…

题目链接: 传送门 Agri-Net Time Limit: 1000MS     Memory Limit: 10000K Description Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course. Farmer…

http://poj.org/problem?id=1125 #include <iostream> #include <cstring> using namespace std; int d[101][101];// dag ATTENTION int num[101];//the number of contracts int edge[101][101];// adjecent edge table int n;//always represent the maxnum of…

http://poj.org/problem?id=1860 #include <cstdio> //#include <queue> //#include <deque> #include <cstring> using namespace std; #define MAXM 202 #define MAXN 101 int n,m; int first[MAXN]; int next[MAXM]; int pto[MAXM]; double earn[M…

http://poj.org/problem?id=1979 #include <cstdio> #include <cstring> using namespace std; const int maxn = 21; bool vis[maxn][maxn]; char maz[maxn][maxn]; int n,m; const int dx[4] = {1,-1,0,0}; const int dy[4] = {0,0,1,-1}; int ans; bool in(int…

题目 http://poj.org/problem?id=1837 题意 单组数据,有一根杠杆,有R个钩子,其位置hi为整数且属于[-15,15],有C个重物,其质量wi为整数且属于[1,25],重物与重物之间,钩子与钩子之间彼此不同.忽略杠杆及重心的影响,有多少种方式使得全部重物都挂上钩子(某些钩子可能挂若干个重物)后杠杆平衡? 思路 由于状态比较小,即使n的五次方也足以承受,而且任意时刻杠杆的状态在[-15 * 25 * 20, 15 * 25 * 20]之间,所以可以直接穷举状态. 感想…