UVa1161 Objective: Berlin（最大流）
题目
Source
https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3602
Description
The administration of a well-known football team has made a study about the lack of support in international away games. This study has concluded that the only reason for this lack of support is the difficulty in organizing the travel arrangements. To help solving this problem, the administration has asked you to build a program that computes the maximum number of people that can fly from a departure city to a destination city, using the available places in regular flights in a given day, and arriving at or before a given time. When traveling from one city to another, a person may make multiple transfers. Each transfer is, at least, 30 minutes long, i.e., the departure time should be, at least 30 minutes after the arrival time. Given a set of flights for a single day and the latest arrival time, your program should compute the maximum number of persons that can fly, directly or indirectly, from departure city to a destination city, arriving at or before the latest arrival time.
Input
The input will contain several test cases, each of them as described below. Consecutive test cases are separated by a single blank line. The first line contains an integer (smaller or equal to 150) indicating the number of cities that have flight connections. The second line contains a string indicating the city of departure. The third line contains a string indicating the destination city. The fourth line contains the latest arrival time, in the format HHMM, where HH is the hour in the day (from 00 to 23) and MM is the minute in the hour (from 00 to 59). The fifth line contains an integer N (smaller or equal to 5000), with the number of existing flights. Each of the following N lines contains the info for each flight. Each such line contains two strings and three integers, separated by blank spaces, O E C D A, where O and E are, respectively, the origin and destination of a flight, C is the number of available places in the flight (from 0 to 300), and D and A are the departure and arrival times in the previously defined format HHMM. All flights start and end in the same day. City names may have up to 8 characters.
Output
For each test case, the output must follow the description below.
The output consists of one single line with an integer stating the maximum number of people that can fly from the origin to the destination city, using the given flights and arriving at or before the given latest arrival time.
Sample Input
4
lisbon
berlin
1500
9
lisbon london 6 1000 1100
london lisbon 6 1130 1230
lisbon paris 5 1000 1100
paris lisbon 4 1130 1230
london paris 1 1130 1300
london berlin 2 1340 1510
berlin london 2 1300 1430
paris berlin 10 1330 1500
berlin paris 9 1300 1430
Sample Output
6
分析
题目大概说有n个城市和m条航线，每条航线都是从某一时刻某一城市出发在某一时刻到达某一城市，且各条航线都有最多人数的限制。现在已知出发城市和目的地城市以及一个最终的时刻，问最多有多少人能在最终时刻前到达目的地。
这题自然会去想把城市看成点，航线看成边。这样的话城市需要按照时间拆点24*60*100=144000，太多了。。然而事实上有连边的点最多只会有2*m个，即10000个，其他10多W都是摆设，所以其实也OK吧。
这题的正解是：
- 把航线看成点，并且拆成两点x和x'，x表示航线开始，另一点x'表示航线结束；
- 新建源点汇点，源点向所有从出发城市是起始城市的航线x连容量INF的边，所有到达城市是目的地且到达时间在最终时刻之前的航线x'向汇点连容量INF的边；
- 所有航线，x向x'连容量为航线限制人数的边，这样就表示航线从开始到结束最多能流过的人数；
- 对于所有的A航线和B航线，满足A航线到达城市为B航线出发城市 且 A航线到达时刻+30分钟在B航线开始时刻之前，由A'到B连容量INF的边，这样就表示A航线结束后可以起身开始进行B航线；
- 跑最大流就是结果了。
其实，这个解法也能看成是把城市按时间拆点并离散化。
不过没想到真的是10000个点去跑网络流。。另外，思维不要局限，图中的点和边可以是各种东西。。要大胆建图跑网络流。。
代码
#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
#include<queue>
#include<algorithm>
using namespace std;
#define INF (1<<30)
#define MAXN 11111
#define MAXM 2222222
struct Edge{
int v,cap,flow,next;
}edge[MAXM];
int vs,vt,NE,NV;
int head[MAXN];
void addEdge(int u,int v,int cap){
edge[NE].v=v; edge[NE].cap=cap; edge[NE].flow=0;
edge[NE].next=head[u]; head[u]=NE++;
edge[NE].v=u; edge[NE].cap=0; edge[NE].flow=0;
edge[NE].next=head[v]; head[v]=NE++;
}
int level[MAXN];
int gap[MAXN];
void bfs(){
memset(level,-1,sizeof(level));
memset(gap,0,sizeof(gap));
level[vt]=0;
gap[level[vt]]++;
queue<int> que;
que.push(vt);
while(!que.empty()){
int u=que.front(); que.pop();
for(int i=head[u]; i!=-1; i=edge[i].next){
int v=edge[i].v;
if(level[v]!=-1) continue;
level[v]=level[u]+1;
gap[level[v]]++;
que.push(v);
}
}
}
int pre[MAXN];
int cur[MAXN];
int ISAP(){
bfs();
memset(pre,-1,sizeof(pre));
memcpy(cur,head,sizeof(head));
int u=pre[vs]=vs,flow=0,aug=INF;
gap[0]=NV;
while(level[vs]<NV){
bool flag=false;
for(int &i=cur[u]; i!=-1; i=edge[i].next){
int v=edge[i].v;
if(edge[i].cap!=edge[i].flow && level[u]==level[v]+1){
flag=true;
pre[v]=u;
u=v;
//aug=(aug==-1?edge[i].cap:min(aug,edge[i].cap));
aug=min(aug,edge[i].cap-edge[i].flow);
if(v==vt){
flow+=aug;
for(u=pre[v]; v!=vs; v=u,u=pre[u]){
edge[cur[u]].flow+=aug;
edge[cur[u]^1].flow-=aug;
}
//aug=-1;
aug=INF;
}
break;
}
}
if(flag) continue;
int minlevel=NV;
for(int i=head[u]; i!=-1; i=edge[i].next){
int v=edge[i].v;
if(edge[i].cap!=edge[i].flow && level[v]<minlevel){
minlevel=level[v];
cur[u]=i;
}
}
if(--gap[level[u]]==0) break;
level[u]=minlevel+1;
gap[level[u]]++;
u=pre[u];
}
return flow;
}
int covert_time(int t){
return t/100*60+t%100;
}
string city[111];
string from[5555],to[5555];
int cap[5555],s[5555],t[5555];
int main(){
int n,m;
while(cin>>n){
string s_city,t_city;
cin>>s_city>>t_city;
int deadline; cin>>deadline;
deadline=covert_time(deadline);
cin>>m;
for(int i=1; i<=m; ++i){
cin>>from[i]>>to[i]>>cap[i]>>s[i]>>t[i];
s[i]=covert_time(s[i]);
t[i]=covert_time(t[i]);
}
vs=0; vt=m*2+1; NV=vt+1; NE=0;
memset(head,-1,sizeof(head));
for(int i=1; i<=m; ++i){
if(from[i]==s_city) addEdge(vs,i,INF);
if(to[i]==t_city && t[i]<=deadline) addEdge(i+m,vt,INF);
addEdge(i,i+m,cap[i]);
}
for(int i=1; i<=m; ++i){
for(int j=1; j<=m; ++j){
if(i==j) continue;
if(to[i]==from[j] && t[i]+30<=s[j]) addEdge(i+m,j,INF);
}
}
printf("%d\n",ISAP());
}
return 0;
}
