poj1797 - Heavy Transportation

                                   想看更多的解题报告:http://blog.csdn.net/wangjian8006/article/details/7870410
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题目大意：有n个城市，m条道路，在每条道路上有一个承载量，现在要求从1到n城市最大承载量，而最大承载量就是从城市1到城市n所有通路上的最大承载量
解题思路：其实这个求最大边可以近似于求最短路，只要修改下找最短路更新的条件就可以了

 

/*4128K 375MSDijkstra邻接矩阵*/#include <iostream>using namespace std;#define MAXV 1010#define min(a,b) (a<b?a:b)int map[MAXV][MAXV],n,m;int dijkstra(){int vis[MAXV],d[MAXV],i,j,v;for(i=1;i<=n;i++){vis[i]=0;d[i]=map[1][i];//这个时候d不代表从1到n的最短路径，而是最大承载量}for(i=1;i<=n;i++){int f=-1;for(j=1;j<=n;j++)if(!vis[j] && d[j]>f){f=d[j];v=j;}vis[v]=1;for(j=1;j<=n;j++)if(!vis[j] && d[j]<min(d[v],map[v][j])){d[j]=min(d[v],map[v][j]);}}return d[n];}int main(){int t,i,j,sum,a,b,c;scanf("%d",&sum);for(t=1;t<=sum;t++){scanf("%d%d",&n,&m);for(i=0;i<=n;i++)for(j=0;j<=n;j++)map[i][j]=0;for(i=1;i<=m;i++){scanf("%d%d%d",&a,&b,&c);map[a][b]=map[b][a]=c;}printf("Scenario #%d:\n",t);printf("%d\n\n",dijkstra());}return 0;}

 

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/*spfa邻接矩阵4156K 469MS */#include <iostream>#include <queue>using namespace std;#define MAXV 1010#define min(a,b) (a<b?a:b)int map[MAXV][MAXV],n,m;int spfa(){queue <int>q;int i,j,v;int vis[MAXV],d[MAXV];for(i=1;i<=n;i++){vis[i]=0;d[i]=0;}q.push(1);vis[1]=1;while(!q.empty()){v=q.front();q.pop();vis[v]=0;for(i=1;i<=n;i++){if(v==1 && map[v][i]){d[i]=map[v][i];q.push(i);vis[i]=1;continue;}if(d[i]<min(d[v],map[v][i])){d[i]=min(d[v],map[v][i]);if(!vis[i]){vis[i]=1;q.push(i);}}}}return d[n];}int main(){int t,i,j,sum,a,b,c;scanf("%d",&sum);for(t=1;t<=sum;t++){scanf("%d%d",&n,&m);for(i=0;i<=n;i++)for(j=0;j<=n;j++)map[i][j]=0;for(i=1;i<=m;i++){scanf("%d%d%d",&a,&b,&c);map[a][b]=map[b][a]=c;}printf("Scenario #%d:\n",t);printf("%d\n\n",spfa());}return 0;}

 

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/*bellman-ford邻接矩阵Time Limit Exceeded*/#include <iostream>using namespace std;#define MAXV 1010#define min(a,b) (a<b?a:b)int map[MAXV][MAXV],n,m;int bellman_ford(){int i,j,v,k;int vis[MAXV],d[MAXV];for(i=1;i<=n;i++) d[i]=map[1][i];for(i=1;i<=n;i++){for(j=1;j<=n;j++){for(k=1;k<=n;k++){if (d[k]<min(d[j],map[j][k]) && map[j][k])  d[k]=min(d[j],map[j][k]);if (d[j]<min(d[k],map[k][j]) && map[k][j])  d[j]=min(d[k],map[k][j]);}}}return d[n];}int main(){int t,i,j,sum,a,b,c;scanf("%d",&sum);for(t=1;t<=sum;t++){scanf("%d%d",&n,&m);for(i=0;i<=n;i++)for(j=0;j<=n;j++)map[i][j]=0;for(i=1;i<=m;i++){scanf("%d%d%d",&a,&b,&c);map[a][b]=map[b][a]=c;}printf("Scenario #%d:\n",t);printf("%d\n\n",bellman_ford());}return 0;}

 

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/*760K 1532MSbellman_ford邻接表*/#include <iostream>using namespace std;#define MAXV 1010#define MAXE 1000010#define min(a,b) (a<b?a:b)struct {int s,e,w;}edge[MAXE];int n,m;int bellman_ford(){int i,j,d[MAXV];for(i=1;i<=n;i++) d[i]=0;d[1]=0xffffff;for (i=1;i<n;i++){for (j=1;j<=m;j++){if (d[edge[j].e]<min(d[edge[j].s],edge[j].w))  d[edge[j].e]=min(d[edge[j].s],edge[j].w);if (d[edge[j].s]<min(d[edge[j].e],edge[j].w))  d[edge[j].s]=min(d[edge[j].e],edge[j].w); }  }return d[n];}int main(){int t,i,sum;scanf("%d",&sum);for(t=1;t<=sum;t++){scanf("%d%d",&n,&m);for(i=1;i<=m;i++){scanf("%d%d%d",&edge[i].s,&edge[i].e,&edge[i].w);}printf("Scenario #%d:\n",t);printf("%d\n\n",bellman_ford());}return 0;}

 

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/*760K 250MSbell_ford邻接表优化*/#include <iostream>using namespace std;#define MAXV 1010#define MAXE 1000010#define min(a,b) (a<b?a:b)struct {int s,e,w;}edge[MAXE];int n,m;int bellman_ford(){int i,j,d[MAXV];for(i=1;i<=n;i++) d[i]=0;d[1]=0xffffff;int flag=1;while(flag){flag=0;for (j=1;j<=m;j++){if (d[edge[j].e]<min(d[edge[j].s],edge[j].w))  {d[edge[j].e]=min(d[edge[j].s],edge[j].w);flag=1;}if (d[edge[j].s]<min(d[edge[j].e],edge[j].w))  {d[edge[j].s]=min(d[edge[j].e],edge[j].w);flag=1;}}  }return d[n];}int main(){int t,i,sum;scanf("%d",&sum);for(t=1;t<=sum;t++){scanf("%d%d",&n,&m);for(i=1;i<=m;i++){scanf("%d%d%d",&edge[i].s,&edge[i].e,&edge[i].w);}printf("Scenario #%d:\n",t);printf("%d\n\n",bellman_ford());}return 0;}

 

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/*bellman-ford邻接矩阵优化4124K 1485MS */#include <iostream>using namespace std;#define MAXV 1010#define min(a,b) (a<b?a:b)int map[MAXV][MAXV],n,m;int bellman_ford(){int i,j,v,k;int vis[MAXV],d[MAXV];for(i=1;i<=n;i++) d[i]=map[1][i];int flag=1;while(flag){flag=0;for(j=1;j<=n;j++){for(k=1;k<=n;k++){if (d[k]<min(d[j],map[j][k]) && map[j][k])  {d[k]=min(d[j],map[j][k]);flag=1;}if (d[j]<min(d[k],map[k][j]) && map[k][j])  {d[j]=min(d[k],map[k][j]);flag=1;}}}}return d[n];}int main(){int t,i,j,sum,a,b,c;scanf("%d",&sum);for(t=1;t<=sum;t++){scanf("%d%d",&n,&m);for(i=0;i<=n;i++)for(j=0;j<=n;j++)map[i][j]=0;for(i=1;i<=m;i++){scanf("%d%d%d",&a,&b,&c);map[a][b]=map[b][a]=c;}printf("Scenario #%d:\n",t);printf("%d\n\n",bellman_ford());}return 0;}