想看更多的解题报告: http://blog.csdn.net/wangjian8006/article/details/7870410
转载请注明出处:http://blog.csdn.net/wangjian8006
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 375MS Dijkstra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 1532MS bellman_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 250MS bell_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; }