problem

1003 Emergency (25)(25 point(s))
As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are marked on the map. When there is an emergency call to you from some other city, your job is to lead your men to the place as quickly as possible, and at the mean time, call up as many hands on the way as possible. Input Each input file contains one test case. For each test case, the first line contains 4 positive integers: N (<= 500) - the number of cities (and the cities are numbered from 0 to N-1), M - the number of roads, C1 and C2 - the cities that you are currently in and that you must save, respectively. The next line contains N integers, where the i-th integer is the number of rescue teams in the i-th city. Then M lines follow, each describes a road with three integers c1, c2 and L, which are the pair of cities connected by a road and the length of that road, respectively. It is guaranteed that there exists at least one path from C1 to C2. Output For each test case, print in one line two numbers: the number of different shortest paths between C1 and C2, and the maximum amount of rescue teams you can possibly gather.\ All the numbers in a line must be separated by exactly one space, and there is no extra space allowed at the end of a line. Sample Input 5 6 0 2
1 2 1 5 3
0 1 1
0 2 2
0 3 1
1 2 1
2 4 1
3 4 1 Sample Output 2 4

anwser

Dijkstra 解法
#include<bits/stdc++.h> #define INF 0x3f3f3f3f
#define Max 511 int N, M, C1, C2;
int Rescue[Max], Map[Max][Max], Dis[Max], Pre[Max], W[Max], Diff[Max];
bool Vis[Max] = {false}; void Dijkstra(int s){
memset(Dis, INF, sizeof(Dis));
memset(W, 0, sizeof(W));
memset(Diff, 0, sizeof(Diff)); Dis[s] = 0;
W[s] = Rescue[s];
Diff[s] = 1;
for(int i = 0; i < N; i++) Pre[i] = i; for(int i = 0; i < N; i++){
int u = 0, minn = INF;
for(int j = 0; j < N; j++){
if(!Vis[j] && Dis[j] < minn){
u = j;
minn = Dis[j];
}
} if(u == C2 || minn == INF) return;
Vis[u] = true; for(int v = 0; v < N; v++) {
if(!Vis[v]) {
if(Dis[u] + Map[u][v] < Dis[v]){
Dis[v] = Dis[u] + Map[u][v];
// Pre[v] = u;
// }
W[v] = W[u] + Rescue[v];
Diff[v] = Diff[u];
}else if (Dis[u] + Map[u][v] == Dis[v]){
Diff[v] += Diff[u];
if(W[u] + Rescue[v] > W[v]){
W[v] = W[u] + Rescue[v];
// Pre[v] = u;
}
} }
}
}
} int main(){
// freopen("test.txt", "r", stdin); memset(Map, INF, sizeof(Map)); std::cin>>N>>M>>C1>>C2;
for(int i = 0; i < N; i++){
std::cin>>Rescue[i];
} for(int i = 0; i < M; i++){
int c1, c2, L;
std::cin>>c1>>c2>>L;
Map[c1][c2] = Map[c2][c1] = L;
} Dijkstra(C1); std::cout<<Diff[C2]<<" "<<W[C2]; return 0;
} /*
5 6 0 2
1 2 1 5 3
0 1 1
0 2 2
0 3 1
1 2 1
2 4 1
3 4 1
*/
DFS解法

#include<bits/stdc++.h>
#include<vector> #define INF 0x3f3f3f3f
#define Max 511 int N, M, C1, C2;
int Rescue[Max], Map[Max][Max], Dis[Max], Pre[Max], W[Max], Diff[Max];
bool Vis[Max] = {false}; void Dijkstra(int s){
memset(Dis, INF, sizeof(Dis));
memset(W, 0, sizeof(W));
memset(Diff, 0, sizeof(Diff)); Dis[s] = 0;
W[s] = Rescue[s];
Diff[s] = 1;
for(int i = 0; i < N; i++) Pre[i] = i; for(int i = 0; i < N; i++){
int u = 0, minn = INF;
for(int j = 0; j < N; j++){
if(!Vis[j] && Dis[j] < minn){
u = j;
minn = Dis[j];
}
} if(u == C2 || minn == INF) return;
Vis[u] = true; for(int v = 0; v < N; v++) {
if(!Vis[v]) {
if(Dis[u] + Map[u][v] < Dis[v]){
Dis[v] = Dis[u] + Map[u][v];
W[v] = W[u] + Rescue[v];
Diff[v] = Diff[u];
}else if (Dis[u] + Map[u][v] == Dis[v]){
Diff[v] += Diff[u];
if(W[u] + Rescue[v] > W[v]){
W[v] = W[u] + Rescue[v];
}
} }
}
}
} int minDis = INF, diff = 0, maxTeam = 0, vis[Max]; void DFS(int v, int dis, int team){
if(v == C2){
if(dis < minDis)
{
minDis = dis;
diff = 1;
maxTeam = team;
}else if(dis == minDis){
diff++;
if(team > maxTeam) maxTeam = team;
}
// std::cout<<team<<std::endl;
return ;
}
vis[v] = 1;
for(int i = 0; i < N; i++)
if(vis[i] == 0 && Map[v][i] != INF)
DFS(i, dis + Map[v][i], team + Rescue[i]);
vis[v] = 0;
} int main(){
// freopen("test.txt", "r", stdin); memset(Map, INF, sizeof(Map));
memset(vis, 0, sizeof(vis)); std::cin>>N>>M>>C1>>C2;
for(int i = 0; i < N; i++){
std::cin>>Rescue[i];
} for(int i = 0; i < M; i++){
int c1, c2, L;
std::cin>>c1>>c2>>L;
Map[c1][c2] = Map[c2][c1] = L;
} // Dijkstra(C1);
// std::cout<<Diff[C2]<<" "<<W[C2]; DFS(C1, 0, Rescue[C1]);
std::cout<<diff<<" "<<maxTeam; return 0;
} /*
5 6 0 2
1 2 1 5 3
0 1 1
0 2 2
0 3 1
1 2 1
2 4 1
3 4 1
*/

experience

  • 注意审题,求的不是最短路,是最短路的不同路条数。
  • 这个图不是单向图,是双向图。
  • Dijkstra算法以及其变种需要熟悉。

    单词复习
  • scattered 分散的

1003 Emergency (25)(25 point(s))的更多相关文章

  1. MySQL5.7.25(解压版)Windows下详细的安装过程

    大家好,我是浅墨竹染,以下是MySQL5.7.25(解压版)Windows下详细的安装过程 1.首先下载MySQL 推荐去官网上下载MySQL,如果不想找,那么下面就是: Windows32位地址:点 ...

  2. PAT 甲级 1006 Sign In and Sign Out (25)(25 分)

    1006 Sign In and Sign Out (25)(25 分) At the beginning of every day, the first person who signs in th ...

  3. 【PAT】1020 Tree Traversals (25)(25 分)

    1020 Tree Traversals (25)(25 分) Suppose that all the keys in a binary tree are distinct positive int ...

  4. 【PAT】1052 Linked List Sorting (25)(25 分)

    1052 Linked List Sorting (25)(25 分) A linked list consists of a series of structures, which are not ...

  5. 【PAT】1060 Are They Equal (25)(25 分)

    1060 Are They Equal (25)(25 分) If a machine can save only 3 significant digits, the float numbers 12 ...

  6. 【PAT】1032 Sharing (25)(25 分)

    1032 Sharing (25)(25 分) To store English words, one method is to use linked lists and store a word l ...

  7. 【PAT】1015 德才论 (25)(25 分)

    1015 德才论 (25)(25 分) 宋代史学家司马光在<资治通鉴>中有一段著名的“德才论”:“是故才德全尽谓之圣人,才德兼亡谓之愚人,德胜才谓之君子,才胜德谓之小人.凡取人之术,苟不得 ...

  8. 1002 A+B for Polynomials (25)(25 point(s))

    problem 1002 A+B for Polynomials (25)(25 point(s)) This time, you are supposed to find A+B where A a ...

  9. PAT 甲级 1010 Radix (25)(25 分)进制匹配(听说要用二分,历经坎坷,终于AC)

    1010 Radix (25)(25 分) Given a pair of positive integers, for example, 6 and 110, can this equation 6 ...

随机推荐

  1. 一次非线上iowait高的情况的检查

    一.现象 iowait高达30%.使用iotop查知jbd2/sda6-8占用60%的io写入.mongodb每秒写入达400k. 必然复现 二.排查 1.先检查是不是mongodb引起的 将mong ...

  2. C++的各种初始化方式

    C++小实验测试:下面程序中main函数里a.a和b.b的输出值是多少? #include <iostream> struct foo { foo() = default; int a; ...

  3. Strusts2笔记9--防止表单重复提交和注解开发

    防止表单重复提交: 用户可能由于各种原因,对表单进行重复提交.Struts2中使用令牌机制防止表单自动提交.以下引用自北京动力节点:

  4. 【Python学习笔记】使用Python进行T检验

    使用Python进行T检验 所需要用到的第三方库有scipy. 均可以通过pip直接安装. pip install scipy numpy 引入第三方库 from scipy import stats ...

  5. SPI子系统分析之二:数据结构【转】

    转自:http://www.cnblogs.com/jason-lu/articles/3164901.html 内核版本:3.9.5 spi_master struct spi_master用来描述 ...

  6. ASP.NET MVC 5使用Filter过滤Action参数防止sql注入,让你代码安全简洁

    在开发程序的过程中,稍微不注意就会隐含有sql注入的危险.今天我就来说下,ASP.NET mvc 5使用Filter过滤Action参数防止sql注入,让你代码安全简洁.不用每下地方对参数的值都进行检 ...

  7. nodejs 接收上传的图片

    1.nodejs接收上传的图片主要是使用formidable模块,服务器是使用的express搭建. 引入formidable var formidable = require('./node_mod ...

  8. Java线程:新特征-有返回值的线程《转》

      原始文章   在Java5之前,线程是没有返回值的,常常为了“有”返回值,破费周折,而且代码很不好写.或者干脆绕过这道坎,走别的路了.   现在Java终于有可返回值的任务(也可以叫做线程)了. ...

  9. Linux入门(一)root密码设置和用户切换

    从这学期开始,本人将会亲自开一个Linux 专题学习包括Linux 常用命令,常见问题的一些解决方法,以及Linux系统下C和C++一些学习经验 下面这张图片是首次安装Ubuntu后第一次设置root ...

  10. Ubuntu 搭建etcd

    一.简介 etcd是一个高可用的分布式键值(key-value)数据库.etcd内部采用raft协议作为一致性算法,etcd基于Go语言实现. 提供配置共享和服务发现的系统比较多,其中最为大家熟知的是 ...