hdu 2544 最短路

Posted 2020-10-31 我只有一件白T恤

最短路

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 80757    Accepted Submission(s): 34969

Problem Description

在每年的校赛里，所有进入决赛的同学都会获得一件很漂亮的t-shirt。但是每当我们的工作人员把上百件的衣服从商店运回到赛场的时候，却是非常累的！所以现在他们想要寻找最短的从商店到赛场的路线，你可以帮助他们吗？

 

Input

输入包括多组数据。每组数据第一行是两个整数N、M（N<=100，M<=10000），N表示成都的大街上有几个路口，标号为1的路口是商店所在地，标号为N的路口是赛场所在地，M则表示在成都有几条路。N=M=0表示输入结束。接下来M行，每行包括3个整数A，B，C（1<=A,B<=N,1<=C<=1000）,表示在路口A与路口B之间有一条路，我们的工作人员需要C分钟的时间走过这条路。
输入保证至少存在1条商店到赛场的路线。

 

Output

对于每组输入，输出一行，表示工作人员从商店走到赛场的最短时间

 

Sample Input

2 1 1 2 3 3 3 1 2 5 2 3 5 3 1 2 0 0

 

Sample Output

3 2

 

思路：最短路水题，练习算法。

• 不知道什么名字的算法：

 

Run ID	Submit Time	Judge Status	Pro.ID	Exe.Time	Exe.Memory	Code Len.	Language	Author
24454955	2018-04-16 21:04:36	Accepted	2544	62MS	2024K	808 B	C++	zqiangda

 

#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
using namespace std;
struct Node{
    int s, e, w;
    Node(int a, int b, int c) :s(a), e(b), w(c){}
};
int main()
{
    int n, m;
    long len[105];
    vector<Node>vec;
    while (cin >> n >> m, n+m){
        vec.clear();
        for (int i = 0; i < m; i++){
            int a, b, c;
            cin >> a >> b >> c;
            vec.push_back(Node(a, b, c));
        }
        memset(len, 125, sizeof(len));
        len[1] = 0;

        int flag = 1;
        while (flag){
            flag = 0;
            for (int i = 0; i < vec.size(); i++){
                int s = vec[i].s, e = vec[i].e, w = vec[i].w;

                if (len[e]>len[s] + w){
                    len[e] = len[s] + w;
                    flag = 1;
                }
                if (len[s] > len[e] + w){
                    len[s] = len[e] + w;
                    flag = 1;
                }
            }
        }
        cout << len[n] << endl;
    }
    return 0;
}

 

•  SPFA算法：

Run ID	Submit Time	Judge Status	Pro.ID	Exe.Time	Exe.Memory	Code Len.	Language	Author
24455250	2018-04-16 21:19:34	Accepted	2544	62MS	2068K	974 B	C++	zqiangda

#include<cstdio>
#include<vector>
#include<cstring>
#include<queue>
#include<iostream>
using namespace std;
struct Node{
    int e, w;
    Node(int b, int c) :e(b), w(c){}
};
int n, m;
long len[105], vis[105];
vector<Node>vec[105];
void spfa(int s)
{
    memset(len, 125, sizeof(len));
    memset(vis, 0, sizeof(vis));
    len[s] = 0;
    queue<int>que;
    que.push(s);
    while (!que.empty()){
        int p = que.front(); que.pop();
        vis[p] = 0;
        for (int i = 0; i < vec[p].size(); i++){
            int e = vec[p][i].e, w = vec[p][i].w;
            if (len[e]>len[p] + w){
                len[e] = len[p] + w;
                if (!vis[e]){
                    que.push(e);
                    vis[e] = 1;
                }
            }

        }
    }

}
int main()
{
    
    while (cin >> n >> m, n+m){
        for (int i = 1; i <= n;i++)
            vec[i].clear();
        for (int i = 0; i < m; i++){
            int a, b, c;
            cin >> a >> b >> c;
            vec[a].push_back(Node(b, c));
            vec[b].push_back(Node(a, c));
        }
        
        spfa(1);
        cout << len[n] << endl;
    }
    return 0;
}

• dijkstra算法：

Run ID	Submit Time	Judge Status	Pro.ID	Exe.Time	Exe.Memory	Code Len.	Language	Author
24455452	2018-04-16 21:32:54	Accepted	2544	46MS	2064K	1123 B	C++	zqiangda

#include<cstdio>
#include<vector>
#include<cstring>
#include<climits>
#include<iostream>
using namespace std;
struct Node{
    int e, w;
    Node(int b, int c) :e(b), w(c){}
};
int n, m;
long len[105], vis[105];
vector<Node>vec[105];
void dijkstra(int s)
{
    memset(len, 125, sizeof(len));
    memset(vis, 0, sizeof(vis));
    for (int i = 0; i < vec[s].size(); i++){
        int e = vec[s][i].e, w = vec[s][i].w;
        len[e] = w;
    }
    len[s] = 0; vis[s] = 1;
    
    for (int i = 0; i < n; i++){
        int min = INT_MAX, p = 1;
        for (int j = 1; j <= n; j++){
            if (len[j] < min&&vis[j] == 0){
                min = len[j]; p = j;
            }
        }

        vis[p] = 1;

        for (int j = 0; j < vec[p].size(); j++){
            int e = vec[p][j].e, w = vec[p][j].w;
            if (len[e]>len[p] + w&&vis[e] == 0){
                len[e] = len[p] + w;
            }
        }
    }
}
int main()
{
    
    while (cin >> n >> m, n+m){
        for (int i = 1; i <= n;i++)
            vec[i].clear();
        for (int i = 0; i < m; i++){
            int a, b, c;
            cin >> a >> b >> c;
            vec[a].push_back(Node(b, c));
            vec[b].push_back(Node(a, c));
        }
        dijkstra(1);
        cout << len[n] << endl;
    }
    return 0;
}

 

• Floyd算法：

#include<cstdio>
#include<vector>
#include<cstring>
#include<climits>
#include<iostream>
using namespace std;
int main()
{
    int n, m;
    long map[105][105];
    while (cin >> n >> m, n+m){
        //memset(map, 125, sizeof(map));
        for (int i = 1; i <= n; i++){
            for (int j = 1; j <= n; j++)
                map[i][j] = INT_MAX;
            map[i][i] = 0;
        }
        for (int i = 0; i < m; i++){
            int a, b, c;
            cin >> a >> b >> c;
            map[a][b] = map[b][a] = c;
        }
        for (int k = 1; k <= n; k++)
        for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++){
            if (map[i][k] + map[k][j] < map[i][j])
                map[i][j] = map[i][k] + map[k][j];
        //    printf("%10d\t", map[i][j]);
        }
        cout << map[1][n] << endl;
    }
    return 0;
}

最后一个待调试~