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描述A city is served by a number of fire stations. Some residents have complained that the distance from their houses to the nearest station is too far, so a new station is to be built. You are to choose the location of the fire station so as to reduce the distance to the nearest station from the houses of the disgruntled residents.
The city has up to 500 intersections, connected by road segments of various lengths. No more than 20 road segments intersect at a given intersection. The location of houses and firestations alike are considered to be at intersections (the travel distance from the intersection to the actual building can be discounted). Furthermore, we assume that there is at least one house associated with every intersection. There may be more than one firestation per intersection.
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无奈...自己的程序POJ可以过....理工的过不去...POJ(可以用Floyd...水过)
标程代码:
1 2 #include <iostream> 3 #include <stdio.h> 4 #include <string.h> 5 #include <queue> 6 using namespace std; 7 #define maxN 510 8 #define MAX 0x0fffffff 9 int max(int x,int y){return x>y?x:y;} 10 struct point{ 11 int v,nex,w; 12 }po[maxN*maxN]; 13 int num,N,M,head[maxN],disY[maxN],disX[maxN],key[maxN],flag[maxN]; 14 bool vis[maxN]; 15 void insert(int u,int v,int w) 16 { 17 po[num].v=v; 18 po[num].w=w; 19 po[num].nex=head[u]; 20 head[u]=num++; 21 } 22 void spfa(int soure,int* dis) 23 { 24 memset(vis,false,sizeof(vis)); 25 queue<int>q; 26 q.push(soure); 27 vis[soure]=true;dis[soure]=0; 28 while(!q.empty()) 29 { 30 int u=q.front();q.pop();vis[u]=false; 31 for(int i=head[u];i!=-1;i=po[i].nex) 32 if(dis[po[i].v]>dis[u]+po[i].w){ 33 dis[po[i].v]=dis[u]+po[i].w; 34 if(!vis[po[i].v]){ 35 q.push(po[i].v); 36 vis[po[i].v]=true; 37 } 38 } 39 } 40 } 41 int main() 42 { 43 //freopen("1.txt","r",stdin); 44 char s[30]; 45 while(~scanf("%d%d",&M,&N)) 46 { 47 getchar(); 48 int i,j,u,v,w,k=1,minn=MAX; 49 num=0; 50 memset(head,-1,sizeof(head)); 51 memset(flag,0,sizeof(flag)); 52 for(i=1;i<=M;i++){scanf("%d",&key[i]);getchar();} 53 if(N==1){printf("1\n");continue;} 54 while(gets(s)!=NULL&&strlen(s)){sscanf(s,"%d%d%d",&u,&v,&w);insert(u,v,w);insert(v,u,w);} 55 for(i=0;i<=N;i++)disX[i]=MAX; 56 for(i=1;i<=M;i++)spfa(key[i],disX); 57 for(j=1;j<=N;j++){ 58 int maxx=0; 59 if(disX[j]==0)continue; 60 for(i=1;i<=N;i++)disY[i]=disX[i]; 61 spfa(j,disY); 62 for(i=1;i<=N;i++)maxx=max(maxx,disY[i]); 63 if(minn>maxx||maxx==0){k=j;minn=maxx;} 64 } 65 printf("%d\n",k); 66 } 67 }
POJ:
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cstring> 5 using namespace std; 6 #define maxn 501 7 #define INF 0x3f3f3f3f 8 int n,m,cost[maxn][maxn],dis[maxn],vis[maxn]; 9 int main() 10 { 11 while(~scanf("%d%d",&m,&n)) 12 { 13 int u,v,c; 14 for(int i=1;i<=n;i++) 15 { 16 dis[i]=INF;vis[i]=0; 17 for(int j=1;j<=n;j++) 18 cost[i][j]=i==j?0:INF; 19 } 20 for(int i=1;i<=m;i++) 21 { 22 scanf("%d",&u); 23 dis[u]=0; 24 vis[u]=1; 25 } 26 while(~scanf("%d%d%d",&u,&v,&c)) 27 cost[u][v]=cost[v][u]=c; 28 for(int k=1;k<=n;k++) 29 for(int i=1;i<=n;i++) 30 for(int j=1;j<=n;j++) 31 cost[i][j]=min(cost[i][j],cost[i][k]+cost[k][j]); 32 for(int i=1;i<=n;i++) 33 if(vis[i]) 34 for(int j=1;j<=n;j++) 35 dis[j]=min(dis[j],cost[i][j]); 36 int ans=INF,pos; 37 for(int i=1;i<=n;i++) 38 { 39 int temp=-1; 40 for(int j=1;j<=n;j++) 41 temp=max(temp,min(dis[j],cost[i][j])); 42 if(ans>temp)ans=temp,pos=i; 43 } 44 printf("%d\n",pos); 45 } 46 return 0; 47 }
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原文地址：http://www.cnblogs.com/wangmengmeng/p/5367819.html
