Sicily 1031. Campus 解题报告

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1031_Campus

题目链接：

http://soj.me/1031

题目大意：

给出四个校区的一些地点之间的距离，地点名用字符串来表示，问某两个地点之间的最短路径长度，典型的单源最短路径题目

思路：

单源最短路径问题可以用dijkstra算法实现，这道题比较麻烦的是用字符串来表示地点，我用的处理方法是建立map得到地点名字到序号的映射，对于每个新输入的地点名字，先在map里面查找是否存在，如果不存在就绑定一个新的序号．地点之间的距离用邻接矩阵来存放．

代码：

#include <iostream>
#include <memory.h>
#include <map>
#include <queue>
#include <vector>
using namespace std;

const int INF = 100000;
int distances[300][300];
int shortest[300];
bool included[301];
map<string, int> m;
int numRoads;
int numPlaces;
string start_name, end_name;


void input_and_init();
int find_shortest();
int find_nearest();

int main() {
    int numTestcases;
    cin >> numTestcases;
    while (numTestcases--) {
        input_and_init();

        if(start_name == end_name)
            cout << 0 << endl;
        else if(m.count(start_name) == 0 || m.count(end_name) == 0)
            cout << -1 << endl;
        else
            cout << find_shortest() << endl;
    }
    return 0;
}
void input_and_init(){
    cin >> numRoads;
    numPlaces = 0;
    m.clear();
    for (int i = 0; i < 300; ++i) {
        for (int j = 0; j < 300; ++j) {
            distances[i][j] = (i == j ? 0 : INF);
        }
    }
    memset(included, false, sizeof(included));
    for (int i = 0; i < 300; ++i) {
        shortest[i] = INF;
    }

    string s1, s2;
    int d;
    for (int i = 0; i < numRoads; ++i) {
        cin >> s1 >> s2 >> d;
        if(m.count(s1) == 0)
            m[s1] = ++numPlaces;
        if(m.count(s2) == 0)
            m[s2] = ++numPlaces;
        distances[m[s1]][m[s2]] = distances[m[s2]][m[s1]] = d;
    }
    cin >> start_name >> end_name;
}

int find_shortest(){
    //Post: 如果起点和终点间有连通则返回最短路径长度，否则返回-1
    int start = m[start_name], end = m[end_name];
    shortest[start] = 0;
    for (int i = 0; i < numPlaces; ++i) {
        int cur = find_nearest();
        included[cur] = true;
        for (int j = 1; j <= numPlaces; ++j) {
            if(!included[j]  && shortest[j] > shortest[cur] + distances[cur][j]){
                shortest[j] = shortest[cur] + distances[cur][j];
            }
        }
    }
    if(shortest[end] < INF)
        return shortest[end];
    else
        return -1;
}
int find_nearest(){
    //返回当前离集合S最近的点的下标
    int nearest_index = 0;
    int shortestDistance = INF;
    for (int i = 1; i <= numPlaces; ++i) {
        if(shortest[i] < shortestDistance && !included[i]){
            nearest_index = i;
            shortestDistance = shortest[i];
        }
    }
    return  nearest_index;
}                                 

Sicily 1031. Campus 解题报告

标签：http   io   ar   os   sp   for   on   art   问题   

原文地址：http://www.cnblogs.com/jolin123/p/4122559.html