HDU 2544 最短路 模板题 SPFA Dijkstra

最短路

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 62803    Accepted Submission(s): 27487

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 11 2 33 31 2 52 3 53 1 20 0

 

Sample Output

32

Dijskstra

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int INF = 0x3f3f3f;int map[110][110],dis[110],vis[110];void Dijkstra(int n,int x){int p,min;for(int i = 1; i <= n; i++){dis[i] = map[1][i];vis[i] = 0;}vis[x] = 1;for(int i = 1; i <= n; i++){min = INF;for(int j = 1; j <= n; j++){if(!vis[j] && dis[j] < min){p = j;min = dis[j];}}vis[p] = 1;for(int j = 1; j<=n; j++){if(!vis[j] && dis[p] + map[p][j] < dis[j]){dis[j] = dis[p] + map[p][j];}}}}int main(){int n,m;while(~scanf("%d%d",&n,&m),n+m){for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){map[i][j] = INF;}}int x,y,value;for(int i = 1; i <= m; i++){scanf("%d%d%d",&x,&y,&value);map[x][y] = map[y][x] = value;}Dijkstra(n,1);printf("%d\n",dis[n]);}return 0; } 

spaf

#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespace std;const int INF = 0x3f3f3f;int map[110][110],dis[110];bool vis[110];int n,m;void spaf(int x){for(int i = 1; i <= n; i++){dis[i] = INF;vis[i] = false;}dis[x] = 0;queue<int>q;q.push(x);vis[x] = true;while(!q.empty()){int now = q.front();q.pop();vis[now] = false;for(int i = 1; i <= n; i++){if(dis[i] > dis[now] + map[now][i]){dis[i] = dis[now] + map[now][i];if(!vis[i]){q.push(i);vis[i] = true;}}}}}int main(){while(~scanf("%d%d",&n,&m),n+m){/*for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){map[i][j] = INF;}}*/for(int i = 1; i<=n; i++){for(int j = 1; j<=i; j++){if(i == j){map[i][j] = 0;}elsemap[i][j] = map[j][i] = INF;}}int x,y,value;for(int i = 1; i <= m; i++){scanf("%d%d%d",&x,&y,&value);map[x][y] = map[y][x] = value;}spaf(1);printf("%d\n",dis[n]);}return 0; } 

SPFA +  邻接表

#include <stdio.h>#include <algorithm>#include <string.h>#include <math.h>#include <queue>using namespace std;const int N = 105;const int M = 10005;const int INF = 99999999;struct Edge{    int v,w,next;}edge[M*2];int head[N];bool vis[N];int dis[N];int n,m;void spfa(int s){    queue<int> q;    memset(vis,false,sizeof(vis));    for(int i=1;i<=n;i++)    {        dis[i]=(i==s)?0:INF;    }    q.push(s);    while(!q.empty())    {        int u = q.front();        q.pop();        vis[u] = false;        for(int k = head[u]; k != -1; k = edge[k].next)        {            int v = edge[k].v,w = edge[k].w;            if(dis[u]+w<dis[v])            {                dis[v] = dis[u]+w;                if(!vis[v])                {                    vis[v] = true;                    q.push(v);                }            }        }    }}void addedge(int u,int v,int w,int &k){    edge[k].v = v;    edge[k].w=w;    edge[k].next = head[u];     head[u]=k++;}int main(){    while(scanf("%d%d",&n,&m)!=EOF,(n+m))    {        memset(head,-1,sizeof(head));        int tot=0,u,v,w;        for(int i=0;i<m;i++){            scanf("%d%d%d",&u,&v,&w);            addedge(u,v,w,tot);            addedge(v,u,w,tot);        }        spfa(1);        printf("%d\n",dis[n]);    }    return 0;}