【BZOJ】3432: [Usaco2014 Jan]Cross Country Skiing （bfs+二分）
http://www.lydsy.com/JudgeOnline/problem.php?id=3432
题目说要相互可达，但是只需要从某个点做bfs然后判断其它点是否可达即可。
原因太简单了。。。。。因为它是abs
所以我们二分D，然后判断即可
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr2(a, b, c) for1(i, 1, b) { for1(j, 1, c) cout << a[i][j]; cout << endl; }
#define printarr1(a, b) for1(i, 1, b) cout << a[i]; cout << endl
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }
const int N=505, M=300005, dx[]={-1, 1, 0, 0}, dy[]={0, 0, 1, -1};
int mp[N][N], must[N][N], n, m, front, tail, vis[N][N], X, Y;
struct dat{ int x, y; }q[M];
void bfs(int D) {
for1(i, 1, n) for1(j, 1, m) vis[i][j]=0;
front=tail=0;
q[tail].x=X, q[tail++].y=Y; vis[X][Y]=1;
int x, y;
while(tail!=front) {
dat &t=q[front++]; if(front==M) front=0;
x=t.x, y=t.y;
rep(i, 4) {
int fx=dx[i]+x, fy=dy[i]+y;
if(fx<1 || fy<1 || fx>n || fy>m || vis[fx][fy] || abs(mp[fx][fy]-mp[x][y])>D) continue;
vis[fx][fy]=1;
q[tail].x=fx; q[tail++].y=fy; if(tail==M) tail=0;
}
}
}
bool check(int D) {
bfs(D);
for1(i, 1, n) for1(j, 1, m) if(must[i][j]==1 && !vis[i][j]) return false;
return true;
}
int main() {
int mx=0;
read(n); read(m);
for1(i, 1, n) for1(j, 1, m) read(mp[i][j]), mx=max(mx, mp[i][j]);
for1(i, 1, n) for1(j, 1, m) {
read(must[i][j]); if(must[i][j]==1) X=i, Y=j;
}
int l=0, r=mx;
while(l<=r) {
int mid=(l+r)>>1;
if(check(mid)) r=mid-1;
else l=mid+1;
}
print(r+1);
return 0;
}
Description
The cross-country skiing course at the winter Moolympics is described by an M x N grid of elevations (1 <= M,N <= 500), each elevation being in the range 0 .. 1,000,000,000. Some of the cells in this grid are designated as waypoints for the course. The organizers of the Moolympics want to assign a difficulty rating D to the entire course so that a cow can reach any waypoint from any other waypoint by repeatedly skiing from a cell to an adjacent cell with absolute elevation difference at most D. Two cells are adjacent if one is directly north, south, east, or west of the other. The difficulty rating of the course is the minimum value of D such that all waypoints are mutually reachable in this fashion.
N*M的格子,每个格子都有一个分值v,有的格子一定要经过.两个格子i,j可以互相到达,当且仅当它们有一条边重复(即上下左右方向),且abs(vi-vj)<=D.
Input
* Line 1: The integers M and N.
* Lines 2..1+M: Each of these M lines contains N integer elevations.
* Lines 2+M..1+2M: Each of these M lines contains N values that are either 0 or 1, with 1 indicating a cell that is a waypoint.
Output
* Line 1: The difficulty rating for the course (the minimum value of D such that all waypoints are still reachable from each-other).
Sample Input
20 21 18 99 5
19 22 20 16 26
18 17 40 60 80
1 0 0 0 1
0 0 0 0 0
0 0 0 0 1
INPUT DETAILS: The ski course is described by a 3 x 5 grid of elevations. The upper-left, upper-right, and lower-right cells are designated as waypoints.
Sample Output
OUTPUT DETAILS: If D = 21, the three waypoints are reachable from each-other. If D < 21, then the upper-right waypoint cannot be reached from the other two.
