UVA 10039 Railroads

这道题好吧，一开始便是拓扑排序的想法，搞了好久，试了多组测试数据，没错啊，可是没过。。。作孽啊，竟然忘了拓扑不能处理环，白浪费了一晚上。。。

只好用动态规划了。。

DP【time】【city】表示在time时刻到达city的最迟出发时间，当然，在这个时间不一定到city。

转移方程挺简单，不说你也会。

#include <iostream>
#include <cstdio>
#include <map>
#include <iomanip>
#include <string>
#include <cstring>
#include <algorithm>
using namespace std;

const int MAXN=105;
const int MAXM=120000;
const int inf=90000000;
int head[MAXN];
struct e{
    int u,v;
    int depart,arrival;
    int next;
}edge[MAXM];
int tot,n,m,limit;
int start_city,des_city;
string start,destin;
int timeh[105][2450];

void addedge(int u,int v,int de,int ar){
    edge[tot].u=u;
    edge[tot].v=v;
    edge[tot].depart=de;
    edge[tot].arrival=ar;
    edge[tot].next=head[u];
    head[u]=tot++;
}

void slove(){
    for(int e=head[start_city];e!=-1;e=edge[e].next){
        if(edge[e].depart>=limit){
            timeh[edge[e].v][edge[e].arrival]=edge[e].depart;
        }
    }
    for(int i=0;i<=2400;i++){
        for(int j=1;j<=n;j++){
            if(timeh[j][i]!=-1){
                for(int e=head[j];e!=-1;e=edge[e].next){
                    if(i<=edge[e].depart){
                        timeh[edge[e].v][edge[e].arrival]=max(timeh[edge[e].v][edge[e].arrival],timeh[j][i]);
                    }
                }
            }
        }
    }
    for(int i=0;i<=2400;i++){
        if(timeh[des_city][i]!=-1){
            cout << "Departure " << setw(4) << setfill('0');
            cout << timeh[des_city][i] << " " << start << endl;
            cout << "Arrival   " << setw(4) << setfill('0');
            cout << i << " " << destin << endl;
            return ;
        }
    }
    cout << "No connection" << endl;
}

int main(){
    string station,pre,cur;
    int cas=0;int T,pretime,curtime,u,v,train;
    scanf("%d",&T);
    while(T--){
        cas++; tot=0;
        memset(head,-1,sizeof(head));
        memset(timeh,-1,sizeof(timeh));
        map<string,int>city;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            cin>>station;
            city[station]=i;
        }
        scanf("%d",&train);
        while(train--){
            scanf("%d",&m);
            for(int i=1;i<=m;i++){
                cin>>curtime>>cur;
                if(i>1){
                    u=city[pre];
                    v=city[cur];
                    addedge(u,v,pretime,curtime);
                }
                pre=cur;
                pretime=curtime;
            }
        }
        cin>>limit>>start>>destin;
        start_city=city[start]; des_city=city[destin];
         cout << "Scenario " << cas << endl;
        slove();
        cout << endl;
    }
    return 0;
}