uva 1161 Objective: Berlin (最大流)
2015-08-09 08:36
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uva 1161 Objective: Berlin
题目大意:你要从A地到B地去,并且最晚要在lt之前到达。现在给你m个航班信息,信息包括:起始地点,降落地点,载客上限,起飞时间,降落时间。中途转机要花费半小时的时间。问在lt之前,可以从A地到达B地的最多的游客数量。
解题思路:以航班为节点进行建图。设置超级源点,连向所有起点为A地的航班,容量为INF;设置超级汇点,使所有降落地点为B点且降落时间在lt之前的航班连向超级汇点。每个航班都要拆成两个节点,中间的边容量为该航班的载客上限。如果i航班的降落地点等于j航班的起始地点,且i航班的降落时间+30 <= j航班的起飞时间,则连接i + m结点(拆出来的结点)和j结点。建完图求最大流。
[code]#include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <cstdlib> #include <iostream> #include <map> using namespace std; const int N = 10006; const int M = 1000005; const int INF = 0x3f3f3f3f; struct flight{ int bp, ep, p, bt, et; }G ; map<string, int> mp; typedef long long ll; int n, lt, m, id, s, t, ss, tt; string S, T; int ec, head , first , que , lev ; int Next[M], to[M], v[M]; void init() { mp.clear(); ec = 0; id = 1; memset(first, -1, sizeof(first)); } void addEdge(int a,int b,int c) { to[ec] = b; v[ec] = c; Next[ec] = first[a]; first[a] = ec++; to[ec] = a; v[ec] = 0; Next[ec] = first[b]; first[b] = ec++; } int BFS() { int kid, now, f = 0, r = 1, i; memset(lev, 0, sizeof(lev)); que[0] = ss, lev[ss] = 1; while (f < r) { now = que[f++]; for (i = first[now]; i != -1; i = Next[i]) { kid = to[i]; if (!lev[kid] && v[i]) { lev[kid] = lev[now] + 1; if (kid == tt) return 1; que[r++] = kid; } } } return 0; } int DFS(int now, int sum) { int kid, flow, rt = 0; if (now == tt) return sum; for (int i = head[now]; i != -1 && rt < sum; i = Next[i]) { head[now] = i; kid = to[i]; if (lev[kid] == lev[now] + 1 && v[i]) { flow = DFS(kid, min(sum - rt, v[i])); if (flow) { v[i] -= flow; v[i^1] += flow; rt += flow; } else lev[kid] = -1; } } return rt; } int dinic() { int ans = 0; while (BFS()) { for (int i = 0; i <= tt; i++) { head[i] = first[i]; } ans += DFS(ss, INF); } return ans; } int getTime(string str) { int a = (str[0] - '0') * 10 + (str[1] - '0'); int b = (str[2] - '0') * 10 + (str[3] - '0'); return a * 60 + b; } int input() { cin >> S >> T; string lit; cin >> lit; lt = getTime(lit); scanf("%d", &m); string a, b; for (int i = 1; i <= m; i++) { cin >> a >> b; if (!mp.count(a)) mp[a] = id++; if (!mp.count(b)) mp[b] = id++; G[i].bp = mp[a], G[i].ep = mp[b]; scanf("%d", &G[i].p); cin >> a >> b; G[i].bt = getTime(a); G[i].et = getTime(b); } if (!mp.count(S) || !mp.count(T)) return 0; s = mp[S], t = mp[T]; ss = 0; tt = m * 2 + 1; return 1; } void build() { for (int i = 1; i <= m; i++) { flight f = G[i]; if (f.bp == s) { addEdge(ss, i, INF); } if (f.ep == t && f.et <= lt) { addEdge(i + m, tt, INF); } addEdge(i, i + m, f.p); for (int j = 1; j <= m; j++) { if (i == j) continue; if (f.ep != G[j].bp) continue; if (f.et + 30 > G[j].bt) continue; addEdge(i + m, j, INF); } } } int main() { while (scanf("%d\n", &n) != EOF) { init(); int flag = input(); if (!flag) { printf("0\n"); continue; } build(); printf("%d\n", dinic()); } return 0; }
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