题意:给定一些航班,每个航班有人数,和起始终止时间,每次转机要花半小时,问限制时间内最多能有多少人从起始城市到终点城市。
析:差不多是裸板网络流的最大流问题,把每个航班都拆成两个点,这两个点之间连接一条流量为这个航班的容量,然后再暴力去查看能不能连接,如果能,
那么就连接一条容量无限的边,然后在源点和汇点加一个无限的容量边。
#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 <unordered_map> //#include <tr1/unordered_map> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; //using namespace std :: tr1; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 10005; const LL mod = 10000000000007; const int N = 1e6 + 5; const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1}; const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1}; const int hr[]= {-2, -2, -1, -1, 1, 1, 2, 2}; const int hc[]= {-1, 1, -2, 2, -2, 2, -1, 1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); } 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 int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } const int maxm = 1000005; struct Edge{ int v, f, next; }; int src, sink; int g[maxn]; int nume; Edge e[maxm]; map<string, int> mp; int cnt; int getid(const string &s){ return mp.count(s) ? mp[s] : mp[s] = cnt++; } void add(int u, int v, int c){ e[++nume].v = v; e[nume].f = c; e[nume].next = g[u]; g[u] = nume; e[++nume].v = u; e[nume].f = 0; e[nume].next = g[v]; g[v] = nume; } queue<int> q; bool vis[maxn + 10]; int d[maxn + 10]; void bfs(){ memset(d, 0, sizeof d); while(!q.empty()) q.pop(); vis[src] = true; q.push(src); while(!q.empty()){ int u = q.front(); q.pop(); for(int i = g[u]; i; i = e[i].next) if(e[i].f && !vis[e[i].v]){ q.push(e[i].v); d[e[i].v] = d[u] + 1; vis[e[i].v] = true; } } } int dfs(int u, int del){ if(u == sink) return del; int ans = 0; for(int i = g[u]; i && del; i = e[i].next) if(e[i].f && d[e[i].v] == d[u] + 1){ int dd = dfs(e[i].v, Min(e[i].f, del)); e[i].f -= dd; e[i^1].f += dd; del -= dd; ans += dd; } return ans; } int maxflow(){ int ans = 0; while(true){ memset(vis, 0, sizeof vis); bfs(); if(!vis[sink]) return ans; ans += dfs(src, INF); } } char s[105], t[105]; struct node{ int u, v, c, start, last; }; node a[maxm]; int cal(char *s){ int ans = 0; ans += ((s[0] - ‘0‘) * 10 + s[1] - ‘0‘) * 60; ans += ((s[2] - ‘0‘) * 10 + s[3] - ‘0‘); return ans; } bool judge(const node &lhs, const node &rhs){ return lhs.v == rhs.u && lhs.last + 30 <= rhs.start; } int main(){ while(scanf("%d", &n) == 1){ scanf("%s", s); cnt = 0; mp.clear(); int ss = getid(s); scanf("%s", t); int tt = getid(t); memset(g, 0, sizeof g); nume = 1; int time; scanf("%s", s); time = cal(s); scanf("%d", &m); char start[10], last[10]; src = 0; sink = 2 * m + 1; for(int i = 1; i <= m; ++i){ scanf("%s %s %d %s %s", s, t, &a[i].c, start, last); a[i].u = getid(s); a[i].v = getid(t); a[i].start = cal(start); a[i].last = cal(last); } for(int i = 1; i <= m; ++i){ if(a[i].u == ss) add(0, i, INF); if(a[i].v == tt && a[i].last <= time) add(i+m, 2*m+1, INF); add(i, i+m, a[i].c); for(int j = 1; j <= m; ++j) if(i != j && judge(a[i], a[j])) add(i+m, j, INF); } printf("%d\n", maxflow()); } return 0; }
时间: 2024-10-14 05:20:39