图论trainning-part-1 A. 最短路

A. 最短路

Time Limit: 1000ms

Memory Limit: 32768KB

64-bit integer IO format: %I64d Java class name: Main

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

Sample Output

3
2

解题：最短路。。。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #define LL long long
13 #define INF 0x3f3f3f
14 using namespace std;
15 int mp[110][110],d[110],n,m;
16 bool vis[110];
17 void dij(){
18     int i,j,theMin,index;
19     for(i = 0; i <= n; i++)
20         d[i] = INF>>1;
21     d[1] = 0;
22     memset(vis,false,sizeof(vis));
23     for(i = 0; i < n; i++){
24         theMin = INF;
25         for(j = 1; j <= n; j++){
26             if(!vis[j] && theMin > d[j]) theMin = d[index = j];
27         }
28         vis[index] = true;
29         if(index == n) break;
30         for(j = 1; j <= n; j++){
31             if(!vis[j] && d[j] > d[index]+mp[index][j])
32                 d[j] = d[index] + mp[index][j];
33         }
34     }
35 }
36 int main(){
37     int i,j,u,v,w;
38     while(scanf("%d%d",&n,&m),n||m){
39         for(i = 0; i <= n; i++)
40         for(j = 0; j <= n; j++)
41             mp[i][j] = INF;
42         for(i = 0; i < m; i++){
43             scanf("%d%d%d",&u,&v,&w);
44             mp[u][v] = mp[v][u] = w;
45         }
46         dij();
47         printf("%d\n",d[n]);
48     }
49     return 0;
50 }

View Code

优先队列版Dijkstra

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <vector>
 6 #include <queue>
 7 #include <climits>
 8 #include <algorithm>
 9 #include <cmath>
10 #define LL long long
11 #define INF 0x3f3f3f
12 using namespace std;
13 #define pii pair<int,int>
14 const int maxn = 102;
15 struct arc {
16     int to,w;
17 };
18 int n,m;
19 vector<arc>g[maxn];
20 bool done[maxn];
21 int d[maxn];
22 priority_queue<pii,vector< pii >,greater< pii > >q;
23 void dj() {
24     int i,j;
25     for(i = 1; i <= n; i++) {
26         done[i] = false;
27         d[i] = INF;
28     }
29     d[1] = 0;
30     while(!q.empty()) q.pop();
31     q.push(make_pair(d[0],1));
32     while(!q.empty()) {
33         pii u =q.top();
34         q.pop();
35         int x = u.second;
36         if(done[x]) continue;
37         done[x] = true;
38         for(i = 0; i < g[x].size(); i++) {
39             j = g[x][i].to;
40             if(d[j] > d[x] + g[x][i].w) {
41                 d[j] = d[x]+g[x][i].w;
42                 q.push(make_pair(d[j],j));
43             }
44         }
45     }
46
47 }
48 int main() {
49     int i,j,u,v,w;
50     while(scanf("%d%d",&n,&m),n||m) {
51         for(i = 1; i <= n; i++)
52             g[i].clear();
53         for(i = 0; i < m; i++) {
54             scanf("%d%d%d",&u,&v,&w);
55             g[u].push_back((arc) {v,w});
56             g[v].push_back((arc) {u,w});
57         }
58         dj();
59         printf("%d\n",d[n]);
60     }
61     return 0;
62 }

View Code

Bellman—Ford算法：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <vector>
 6 #include <queue>
 7 #include <climits>
 8 #include <algorithm>
 9 #include <cmath>
10 #define LL long long
11 #define INF 0x3f3f3f
12 using namespace std;
13 #define pii pair<int,int>
14 const int maxn = 10010;
15 struct arc {
16     int u,v,w;
17 } g[maxn];
18 int n,m,cnt;
19 int d[maxn];
20 void bf() {
21     int i,j;
22     for(i = 1; i <= n; i++)
23         d[i] = INF;
24     d[1] = 0;
25     bool flag;
26     for(i = 1; i < n; i++) {
27         flag = true;
28         for(j = 0; j < cnt; j++)
29             if(d[g[j].v] > d[g[j].u]+g[j].w){
30                  d[g[j].v] = d[g[j].u]+g[j].w;
31                  flag = false;
32             }
33         if(flag) break;
34     }
35 }
36 int main() {
37     int i,u,v,w;
38     while(scanf("%d%d",&n,&m),n||m) {
39         for(cnt = i = 0; i < m; i++) {
40             scanf("%d%d%d",&u,&v,&w);
41             g[cnt++] = (arc) {u,v,w};
42             g[cnt++] = (arc) {v,u,w};
43         }
44         bf();
45         printf("%d\n",d[n]);
46     }
47     return 0;
48 }

View Code

spfa算法版：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <vector>
 6 #include <queue>
 7 #include <climits>
 8 #include <algorithm>
 9 #include <cmath>
10 #define LL long long
11 #define INF 0x3f3f3f
12 using namespace std;
13 #define pii pair<int,int>
14 const int maxn = 1001;
15 struct arc {
16     int to,w;
17 };
18 vector<arc>g[maxn];
19 queue<int>q;
20 int n,m,d[maxn];
21 bool done[maxn];
22 void spfa() {
23     int i,j;
24     for(i = 1; i <= n; i++){
25         d[i] = INF;
26         done[i] = false;
27     }
28     d[1] = 0;
29     while(!q.empty()) q.pop();
30     done[1] = true;
31     q.push(1);
32     while(!q.empty()){
33         int temp = q.front();
34         q.pop();
35         done[temp] = false;
36         for(i = 0; i < g[temp].size(); i++){
37             j = g[temp][i].to;
38             if(d[j] > d[temp]+g[temp][i].w){
39                 d[j] = d[temp]+g[temp][i].w;
40                 if(!done[j]){
41                     q.push(j);
42                     done[j] = true;
43                 }
44             }
45         }
46     }
47
48 }
49 int main() {
50     int i,u,v,w;
51     while(scanf("%d%d",&n,&m),n||m) {
52         for(i = 1; i <= n; i++)
53             g[i].clear();
54         for(i = 0; i < m; i++) {
55             scanf("%d%d%d",&u,&v,&w);
56             g[u].push_back((arc){v,w});
57             g[v].push_back((arc){u,w});
58         }
59         spfa();
60         printf("%d\n",d[n]);
61     }
62     return 0;
63 }

View Code

转载于:https://www.cnblogs.com/crackpotisback/p/3858673.html