标签:include mon const 最小路径覆盖 题意 i+1 back opera stack
题意:有 n 个客人,要从 si 到 ti,每个人有一个出发时间,现在让你安排最少和出租车去接,在接客人时至少要提前一分钟到达客人的出发地点。
析:把每个客人看成一个结点,然后如果用同一个出租车接的话,那么肯定是先接 u 然后再去接 v,也就是有一条边 u->v,画图看的就成知道,这是一个最小路径覆盖的问题。把每个结点拆成 X和 Y 然后如果 u 能连 v,那么就 Xu -> Yv,然后跑一次二分最大匹配,那么答案就是 n - 最大匹配数。
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> #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 pu push_up #define pd push_down #define cl clear() #define all 1,n,1 #define FOR(i,x,n) for(int i = (x); i < (n); ++i) #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 1e20; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1000 + 10; const int maxm = 100 + 10; const int mod = 1000000; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, -1, 0, 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; } struct Edge{ int to, next; }; Edge edge[maxn*maxn]; int head[maxn], cnt; void addEdge(int u, int v){ edge[cnt].to = v; edge[cnt].next = head[u]; head[u] = cnt++; } bool used[maxn]; int match[maxn]; bool dfs(int u){ used[u] = 1; for(int i = head[u]; ~i; i = edge[i].next){ int v = edge[i].to, w = match[v]; if(w < 0 || !used[w] && dfs(w)){ match[u] = v; match[v] = u; return true; } } return false; } struct Pessonger{ int time, sx, sy, tx, ty, dist; bool operator < (const Pessonger &p) const{ return time < p.time; } }; Pessonger pess[maxn]; bool judge(int i, int j){ int t = pess[i].dist + pess[i].time + abs(pess[i].tx-pess[j].sx) + abs(pess[i].ty-pess[j].sy); return t < pess[j].time; } int main(){ int T; cin >> T; while(T--){ scanf("%d", &n); for(int i = 0; i < n; ++i){ int h, s; scanf("%d:%d %d %d %d %d", &h, &s, &pess[i].sx, &pess[i].sy, &pess[i].tx, &pess[i].ty); pess[i].time = h * 60 + s; pess[i].dist = abs(pess[i].sx - pess[i].tx) + abs(pess[i].sy - pess[i].ty); } ms(head, -1); cnt = 0; FOR(i, 0, n) for(int j = i+1; j < n; ++j){ if(judge(i, j)) addEdge(i<<1, j<<1|1); } ms(match, -1); int ans = 0; for(int i = 0; i < (n<<1); ++i) if(match[i] < 0){ ms(used, 0); if(dfs(i)) ++ans; } printf("%d\n", n - ans); } return 0; }
UVaLive 3126 Taxi Cab Scheme (最小路径覆盖)
标签:include mon const 最小路径覆盖 题意 i+1 back opera stack
原文地址:http://www.cnblogs.com/dwtfukgv/p/7624791.html