"Bzoj 1295" [SCOI2009] the longest distance

1295: [SCOI2009] Maximum distance time limit: ten Sec Memory Limit: 162 MB
Submit: 945 Solved: 492
[Submit] [Status] [Discuss] Descriptionwindy has a rectangular land, is divided into n*m block 1*1 small lattice. Some grids contain obstructions. If you can walk from lattice A to lattice B, then the distance of two squares is the Euclidean distance from the center of the two lattice. If from lattice a can not go to lattice B, there is no distance. If lattice x and lattice y have common edges, and x and y do not contain obstructions, you can go from X to Y. If the windy can remove the T block obstacle, the maximum distance between all the lattice is obtained. Ensure that there is at least one lattice that does not contain obstructions after removing the T block barrier. Input file maxlength.in The first line contains three integers, N M T. Next there are n lines, each line a string of length m, ' 0 ' for the empty lattice, ' 1 ' indicates that the lattice contains obstructions. The output file Maxlength.out contains a floating-point number that retains 6 decimal places. Sample Input"Input Sample One"
3 3 0
001
001
110


"Input Sample Two"
4 3 0
001
001
011
000


"Input Sample Three"
3 3 1
001
001
001

Sample Output"Output Example One"
1.414214

"Output Example II"
3.605551

"Output Example three"
2.828427
HINT

20% of the data, meet 1 <= n,m <= 0 <= T <= 0.
40% of the data, meet 1 <= n,m <= 0 <= T <= 2.
100% of the data, meet 1 <= n,m <= 0 <= T <= 30.

Source

Day2

spfa+ violence.

For each lattice, use SPFA to preprocess him to any grid at least to remove a few obstacles.

Then the brute force enumeration needs to remove the lattice of the barrier <=t and output the maximum distance.

#include <iostream> #include <cstring> #include <cstdio> #include <cstdlib> #include <cmath > #include <algorithm> #define M 900000+5#define PA pair<int,int> #define MP Make_pair#include <queue > #define INF 0x3f3f3f3fusing namespace std;struct data{double dis;int need;} R[m];char s[100];int f[5][3],tot=0,d[35][35],inq[35][35],n,m,t,a[35][35];queue<pa> q;double dis[35][35][35][ 35];int C (int x,int y) {return (x-1) *m+y;} bool OK (int x,int y) {if (x<1| | y<1| | x>n| | Y&GT;M) return False;return true;} void Spfa (int x,int y) {for (Int. i=1;i<=n;i++) for (int j=1;j<=m;j++) d[i][j]=inf,inq[i][j]=0;d[x][y]=a[x][y]; Q.push (MP (x, y)); Inq[x][y]=1;while (!q.empty ()) {PA p=q.front (); Q.pop (); int xx=p.first,yy=p.second;inq[xx][yy]=0; for (int i=1;i<=4;i++) {int nx=xx+f[i][1],ny=yy+f[i][2];if (OK (nx,ny) &&d[xx][yy]+a[nx][ny]<d[nx][ny]) {d[nx][ny]=d[xx][yy]+a[nx][ny];if (!inq[nx][ny]) Inq[nx][ny]=1,q.push (MP (Nx,ny));}}} for (int i=1;i<=n;i++) for (inT j=1;j<=m;j++) r[++tot].dis=dis[i][j][x][y],r[tot].need=d[i][j];} BOOL CMP (data A,data b) {return a.dis>b.dis;} int main () {f[1][1]=f[2][1]=0,f[1][2]=1,f[2][2]=-1;f[3][2]=f[4][2]=0,f[3][1]=1,f[4][1]=-1;scanf ("%d%d%d", &n, &m,&t), for (int i=1;i<=n;i++) {scanf ("%s", 1+s), and for (int j=1;j<=m;j++) a[i][j]=s[j]-' 0 ';} for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) for (int x=1;x<=n;x++) for (int y=1;y<=m;y++) dis[i][j][x][y]= sqrt (double) (x-i) * (x-i) + (y-j) * (Y-J)); for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) SPFA (i,j); Sort (R+1,r+1+tot , CMP), for (int i=1;i<=tot;i++) if (r[i].need<=t) {printf ("%.6lf\n", R[i].dis); return 0;} return 0;}

"Bzoj 1295" [SCOI2009] the longest distance