链接:http://acm.hdu.edu.cn/showproblem.php?pid=6331
Walking Plan
Problem Description
There are
n
intersections in Bytetown, connected with m
one way streets. Little Q likes sport walking very much, he plans to walk for q
days. On the i
-th day, Little Q plans to start walking at the s
i![]()
-th intersection, walk through at least k
i![]()
streets and finally return to the t
i![]()
-th intersection.
Little Q's smart phone will record his walking route. Compared to stay healthy, Little Q cares the statistics more. So he wants to minimize the total walking length of each day. Please write a program to help him find the best route.
Little Q's smart phone will record his walking route. Compared to stay healthy, Little Q cares the statistics more. So he wants to minimize the total walking length of each day. Please write a program to help him find the best route.
Input
The first line of the input contains an integer
T(1≤T≤10)
, denoting the number of test cases.
In each test case, there are 2
integers n,m(2≤n≤50,1≤m≤10000)
in the first line, denoting the number of intersections and one way streets.
In the next m
lines, each line contains 3
integers u
i
,v
i
,w
i
(1≤u
i
,v
i
≤n,u
i
≠v
i
,1≤w
i
≤10000)
, denoting a one way street from the intersection u
i![]()
to v
i![]()
, and the length of it is w
i![]()
.
Then in the next line, there is an integer q(1≤q≤100000)
, denoting the number of days.
In the next q
lines, each line contains 3
integers s
i
,t
i
,k
i
(1≤s
i
,t
i
≤n,1≤k
i
≤10000)
, describing the walking plan.
In each test case, there are 2
In the next m
Then in the next line, there is an integer q(1≤q≤100000)
In the next q
Output
For each walking plan, print a single line containing an integer, denoting the minimum total walking length. If there is no solution, please print -1.
Sample Input
2 3 3 1 2 1 2 3 10 3 1 100 3 1 1 1 1 2 1 1 3 1 2 1 1 2 1 1 2 1 1
Sample Output
111 1 11 -1
Source
Recommend
chendu
这题时间复杂度卡的。。。。
题解:这题主要用来分块+DP+Folyd.对于数据范围,我们分100位每一块(一般大一点,我取110 Orz).我们可以先预处理出任意两点间走从0~110步的最短路,然后利用走100为一个单位步,
去更新1*100,2*100,....100*100步的最短路,
由于是至少为K条路的最短路,因此>=k. 我们可以可以再预处理更新一遍恰好走x*100步的情况,查找还有没有于x*100的情况使得i->j的距离变小(因为最多50个点,所以不会超过100) 我们把K 分为K/100,,和K%100,分别求;
参考代码为:
#include<bits/stdc++.h> using namespace std; const int INF=0x3f3f3f3f; const int N=55,M=110,maxn=10010; int T,n,m,q,u,v,w,s,t,K; int a[maxn][N][N],b[maxn][N][N],Map[N][N]; int flag[N][N],dis[N][N]; void pre_work(int x[N][N],int y[N][N],int z[N][N]) { for(int i=0;i<n;i++) { for(int j=0;j<n;j++) { flag[i][j]=INF; for(int k=0;k<n;k++) flag[i][j]=min(flag[i][j],x[i][k]+y[k][j]); } } for(int i=0;i<n;i++) for(int j=0;j<n;j++) z[i][j]=flag[i][j]; } int main() { ios::sync_with_stdio(false); cin.tie(0); cin>>T; while(T--) { cin>>n>>m; for(int i=0;i<n;i++) { for(int j=0;j<n;j++) Map[i][j]=INF; } while(m--) { cin>>u>>v>>w; Map[u-1][v-1]=min(Map[u-1][v-1],w); } for(int i=0;i<n;i++) { for(int j=0;j<n;j++) a[0][i][j]=b[0][i][j]= i==j? 0:INF; } for(int i=1;i<M;i++) pre_work(a[i-1],Map,a[i]);//处理出经过i步从 x->y 的最短路 for(int i=1;i<M;i++) pre_work(b[i-1],a[100],b[i]);//处理出从 x->y 恰好走 100*i步 //Floyd for(int i=0;i<n;i++) { for(int j=0;j<n;j++) dis[i][j]= i==j? 0:Map[i][j]; } for(int k=0;k<n;k++) { for(int i=0;i<n;i++) { for(int j=0;j<n;j++) dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]); } } for(int x=0;x<M;x++) { for(int i=0;i<n;i++) { for(int j=0;j<n;j++) { flag[i][j]=INF; for(int k=0;k<n;k++) flag[i][j]=min(flag[i][j],b[x][i][k]+dis[k][j]); } } for(int i=0;i<n;i++) for(int j=0;j<n;j++) b[x][i][j]=flag[i][j]; } cin>>q; while(q--) { cin>>s>>t>>K; s--,t--; int r=K/100,l=K%100,ans=INF; for(int i=0;i<n;i++) ans=min(ans,b[r][s][i]+a[l][i][t]); if(ans>=INF) cout<<-1<<endl; else cout<<ans<<endl; } } return 0; }