题意:给定 n 个人坐标, m 辆车的坐标,还有人的速度,要求每个人要进一辆不同的车,问你所有都进车的最短时间是多少。
析:首先二分时间 mid,很明显就是最后那个人进车的时间,然后如果把第 i 个人到时第 j 辆车的时间小于 mid,那么就从 i 向 j + n 连一条边,然后进行十分匹配,如果是完全匹配,匹配数等于 n,那么就是可以的,否则就是不可以。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #include <assert.h> #include <bitset> #include <numeric> #define debug() puts("++++") #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define fi first #define se second #define pb push_back #define sqr(x) ((x)*(x)) #define ms(a,b) memset(a, b, sizeof a) #define sz size() #define be begin() #define ed end() #define pu push_up #define pd push_down #define cl clear() #define lowbit(x) -x&x //#define all 1,n,1 #define FOR(i,n,x) for(int i = (x); i < (n); ++i) #define freopenr freopen("in.in", "r", stdin) #define freopenw freopen("out.out", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e17; const double inf = 1e20; const double PI = acos(-1.0); const double eps = 1e-6; const int maxn = 100 + 10; const int maxm = 1e6 + 10; const LL mod = 1000000007; const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1}; const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c) { return r >= 0 && r < n && c >= 0 && c < m; } inline int readInt(){ int x; scanf("%d", &x); return x; } int x[maxn], y[maxn]; int a[maxn], b[maxn]; double v; LL d[maxn][maxn]; LL dist(int i, int j){ return sqr((LL)(a[i]-x[j])) + sqr((LL)(b[i]-y[j])); } int match[maxn<<1]; bool vis[maxn<<1]; vector<int> G[maxn<<1]; void addEdge(int u, int v){ G[u].pb(v); G[v].pb(u); } void init(){ FOR(i, n+m+5, 0) G[i].cl; } bool dfs(int u){ vis[u] = 1; for(int i = 0; i < G[u].sz; ++i){ int v = G[u][i]; int w = match[v]; if(w < 0 || !vis[w] && dfs(w)){ match[u] = v; match[v] = u; return true; } } return false; } bool judge(double mid){ int ans = 0; init(); FOR(i, n, 0) FOR(j, m, 0) if((double)d[i][j] <= sqr(mid * v)) addEdge(i, j + n); ms(match, -1); for(int i = 0; i < n; ++i) if(match[i] < 0){ ms(vis, 0); if(dfs(i)) ++ans; } return ans == n; } int main(){ while(scanf("%d %d", &n, &m) == 2){ for(int i = 0; i < n; ++i) scanf("%d %d", a + i, b + i); for(int i = 0; i < m; ++i){ scanf("%d %d", x + i, y + i); for(int j = 0; j < n; ++j) d[j][i] = dist(j, i); } scanf("%lf", &v); double l = 0., r = 1e7 / v; while(r-l > eps){ double mid = (l + r) / 2.; if(judge(mid)) r = mid; else l = mid; } printf("%.2f ", l); } return 0; }