题目链接. 分析: 最大的敌人果然不是别人,就是她(英语). 每种代表车型的串,他们的distance就是串中不同字符的个数,要求算出所有串的distance's 最小 sum. AC代码如下: #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> using namespace std; ; <<); ]; unsigned short G[max…

Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 18981   Accepted: 7321 Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for brick…

Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 19772   Accepted: 7633 Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for brick…

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…

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…

Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 27335   Accepted: 10634 Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bric…

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 读不懂题再简单也不会做,英语是硬伤到哪都是真理,sad++. 此题就是一个最小生成树,两点之间的权值是毎两串之间的不同字母数. #include<stdio.h> #include<string.h> ; <<; int map[N][N],vis[N],dis[N]; int n,sum; void prim() { int pos; ; i <= n; i ++) { dis[i] = map[][…

 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[]…

题目链接: https://vjudge.net/problem/POJ-1789 题目大意: 用一个7位的string代表一个编号,两个编号之间的distance代表这两个编号之间不同字母的个数.一个编号只能由另一个编号"衍生"出来,代价是这两个编号之间相应的distance,现在要找出一个"衍生"方案,使得总代价最小,也就是distance之和最小. 思路: 最小生成树模板题,这里是稠密图,应该用prim算法 直接在原来模板的基础上稍加改动即可 #include…