bzoj4745: [Usaco2016 Dec]Cow Checklist
Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 1 Solved: 1
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Description
Every day, Farmer John walks through his pasture to check on the well-being of each of his cows. Onh
is farm he has two breeds of cows, Holsteins and Guernseys. His HH Holsteins are conveniently number
ed 1…H, and his GG Guernseys are conveniently numbered 1…G (1≤H≤1000,1≤G≤1000). Each cowis loc
ated at a point in the 2D plane (not necessarily distinct).Farmer John starts his tour at Holstein 1
, and ends at Holstein HH. He wants to visit each cow along the way, and for convenience in maintain
ing his checklist of cows visited so far, he wants to visit the Holsteins and Guernseys in theorder
in which they are numbered. In the sequence of all H+GH+G cows he visits, the Holsteins numbered 1…
H should appear as a (not necessarily contiguous) subsequence, and likewise for the Guernseys. Other
wise stated, the sequence of all H+GH+G cows should be formed by interleaving the list of Holsteins
numbered 1…H with the list of Guernseys numbered 1…GWhen FJ moves from one cow to another cow trav
eling a distance of D, he expends D2 energy. Please help him determine the minimum amount ofenergy r
equired to visit all his cows according to a tour as described above.
Input
The first line of input contains H and G, separated by a space.
The next H lines contain the xx and yy coordinates of the HH Holsteins, and the next G lines after
that contain coordinates of the Guernseys. Each coordinate is an integer in the range 0…1000
Output
Write a single line of output, giving the minimum energy required for FJ‘s tour of all the cows.
Sample Input
3 2
0 0
1 0
2 0
0 3
1 3
Sample Output
20
HINT
Source
f[i][j][0/1]表示第一个数组匹配到i 第二个到j 当前在第几个的最小代价 转移见代码
1 #include<bits/stdc++.h>
2 #define rep(i,l,r) for(int i=l;i<=r;++i)
3 using namespace std;
4 const int N=2015;
5 struct zs{
6 int x,y;
7 }s[N];
8 typedef long long ll;
9 int n,m,mp[N][N];
10 ll f[N][N][2];
11 int main(){
12 scanf("%d%d",&n,&m);
13 rep(i,1,n) scanf("%d%d",&s[i].x,&s[i].y);
14 rep(i,1,m) scanf("%d%d",&s[i+n].x,&s[i+n].y);
15 rep(i,1,n+m) rep(j,1,n+m) mp[i][j]=(s[i].x-s[j].x)*(s[i].x-s[j].x)+(s[i].y-s[j].y)*(s[i].y-s[j].y);
16 rep(i,0,n) rep(j,0,m) rep(k,0,1) f[i][j][k]=1LL<<50;
17 f[1][0][0]=0;
18 rep(i,1,n) rep(j,0,m) {
19 f[i][j][0]=min(f[i][j][0],min(f[i-1][j][0]+mp[i-1][i],f[i-1][j][1]+mp[n+j][i]));
20 if(j)f[i][j][1]=min(f[i][j][1],min(f[i][j-1][0]+mp[i][n+j],f[i][j-1][1]+mp[n+j-1][n+j]));
21 }
22 printf("%lld\n",f[n][m][0]);
23 }