Wednesday, June 29, 2016

CCF 201409-4 最优配餐


http://www.cnblogs.com/Deribs4/p/4368439.html

  栋栋最近开了一家餐饮连锁店,提供外卖服务。随着连锁店越来越多,怎么合理的给客户送餐成为了一个急需解决的问题。   栋栋的连锁店所在的区域可以看成是一个n×n的方格图(如下图所示),方格的格点上的位置上可能包含栋栋的分店(绿色标注)或者客户(蓝色标注),有一些格点是不能经过的(红色标注)。   方格图中的线表示可以行走的道路,相邻两个格点的距离为1。栋栋要送餐必须走可以行走的道路,而且不能经过红色标注的点。
   送餐的主要成本体现在路上所花的时间,每一份餐每走一个单位的距离需要花费1块钱。每个客户的需求都可以由栋栋的任意分店配送,每个分店没有配送总量的限制。   现在你得到了栋栋的客户的需求,请问在最优的送餐方式下,送这些餐需要花费多大的成本。
输入格式
  输入的第一行包含四个整数n, m, k, d,分别表示方格图的大小、栋栋的分店数量、客户的数量,以及不能经过的点的数量。   接下来m行,每行两个整数xi, yi,表示栋栋的一个分店在方格图中的横坐标和纵坐标。   接下来k行,每行三个整数xi, yi, ci,分别表示每个客户在方格图中的横坐标、纵坐标和订餐的量。(注意,可能有多个客户在方格图中的同一个位置)   接下来d行,每行两个整数,分别表示每个不能经过的点的横坐标和纵坐标。
输出格式
  输出一个整数,表示最优送餐方式下所需要花费的成本。
样例输入
10 2 3 3 1 1 8 8 1 5 1 2 3 3 6 7 2 1 2 2 2 6 8
样例输出
29
评测用例规模与约定
  前30%的评测用例满足:1<=n <=20。   前60%的评测用例满足:1<=n<=100。   所有评测用例都满足:1<=n<=1000,1<=m, k, d<=n^2。可能有多个客户在同一个格点上。每个客户的订餐量不超过1000,每个客户所需要的餐都能被送到。


http://blog.csdn.net/fishseeker/article/details/50975384
就是在输入时将所有的卖家入队,之后开始BFS,从第一个卖家开始进行一下广搜,把周围点入队,然后从第二个卖家再开始,然后第三个第四个……感觉巧妙之处就是在于这样相当于每个卖家的点以几乎相等的速度向外扩张,谁先碰到顾客的点,就给顾客送餐。
由店出发找买家:
10 int n,m,k,d;
11 short map[1005][1005],dir[5][2]={{1,0},{0,1},{-1,0},{0,-1}};
12 bool within(int x,int y){
13     if(x<=0||x>n||y<=0||y>n)
14         return false;
15     return true;
16 }
17 struct node{
18     int x,y,step;
19     node(int a=0,int b=0,int c=0){
20         x=a;
21         y=b;
22         step=c;
23     }
24 };
25 queue<node> q;
26 long long bfs(){
27     node n;
28     int t=0;
29     long long sum=0;
30     while(!q.empty()){
31         n=q.front();
32         q.pop();
33         int i,x,y;
34         for(i=0;i<4;i++){
35             x=n.x+dir[i][0];
36             y=n.y+dir[i][1];
37             if(within(x,y)&&!(map[x][y]&1)){
38                 map[x][y]|=1;
39                 if(map[x][y]&2){
40                     sum+=(map[x][y]>>2)*(n.step+1);
41                     t++;
42                     if(t==k)
43                         return sum;
44                 }
45                 q.push(node(x,y,n.step+1));
46             }
47         }
48     }
49 }
50 int main(){
51     //freopen("D:\\INPUT.txt","r", stdin);
52     while(scanf("%d %d %d %d",&n,&m,&k,&d)!=EOF){
53         int i,j,x,y;
54         memset(map,0,sizeof(map));
55         for(i=0;i<m;i++){
56             scanf("%d %d",&x,&y);
57             map[x][y]|=1;
58             q.push(node(x,y,0));
59         }
60         for(i=0;i<k;i++){
61             scanf("%d %d %d",&x,&y,&j);//大于0的表示定餐数目
62             map[x][y]=(map[x][y]|2)+(j<<2);
63         }
64         for(i=0;i<d;i++){
65             scanf("%d %d",&x,&y);
66             map[x][y]|=1;
67         }
68         printf("%I64d\n",bfs());
69     }
70     return 0;
71 }
由买家出发找店家(超时,只拿了60):


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