[bzoj1614]: [Usaco2007 Jan]Telephone Lines架设电话线

Posted 2020-10-02 Vincent_hwh

传送门

题意：给一个图，定义两点间的距离为路径上最大的边权，可以将路径上不多于k条边的权值变为0，求两点间最小距离

二分答案，判断时只要将大于当前二分值的边记为1，否则记为0，做一遍spfa，判断dist是否大于k即可

无向边数忘记乘2卡了半天T_T

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 #define maxn 1010
 6 #define maxe 20100
 7 struct node{
 8     int to,next,weight;
 9 }Edge[maxe];
10 int n,p,k;
11 int cnt=0,mxl=0;
12 int last[maxn],q[maxn*10],dist[maxn];
13 bool inq[maxn];
14 int read(){
15     int x=0,f=1;
16     char ch=getchar();
17     while (ch<‘0‘||ch>‘9‘) {
18         if (ch==‘-‘) f=-1;
19         ch=getchar();
20     }
21     while (ch>=‘0‘&&ch<=‘9‘){
22         x=x*10+ch-‘0‘;
23         ch=getchar();
24     }
25     return x*f;
26 }
27 void insert(int u,int v,int w){
28     Edge[++cnt].to=v;
29     Edge[cnt].next=last[u];
30     last[u]=cnt;
31     Edge[cnt].weight=w;
32 }
33 bool ok(int cost){
34     memset(inq,0,sizeof(inq));
35     memset(dist,127/3,sizeof(dist));
36     memset(q,0,sizeof(q));
37     int head=0,tail=1,now,len,i;
38     inq[1]=1;q[head]=1;dist[1]=0;
39     while (head<tail){
40         now=q[head++];
41         len=0;
42         i=last[now];
43         while (i){
44             len=Edge[i].weight>cost?dist[now]+1:dist[now];
45             if (len<dist[Edge[i].to]){
46                 dist[Edge[i].to]=len;
47                 if (!inq[Edge[i].to]){
48                     q[tail++]=Edge[i].to;
49                     inq[Edge[i].to]=1;
50                 }
51             }
52             i=Edge[i].next;
53         }
54         inq[now]=0;
55     }
56     return dist[n]>k?0:1;
57 }
58 int main(){
59     n=read();p=read();k=read();
60     int u,v,w;
61     memset(last,0,sizeof(last));
62     for (int i=1;i<=p;i++){
63         u=read(),v=read(),w=read();
64         insert(u,v,w);
65         insert(v,u,w);
66         mxl=max(mxl,w);
67     }
68     int l=0,r=mxl,mid;
69     int ans=-1;
70     while (l<=r){
71         mid=(l+r)>>1;
72         if (ok(mid)){
73             ans=mid;r=mid-1;
74         }
75         else l=mid+1;
76     }
77     printf("%d\n",ans);
78     return 0;
79 }

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