UVa12661: Funny Car Racing 题解

这题仍然可以跑dijkstra，只是在通过边的时候边的权值不再是通过时间，还要通过数学计算加上一些等待时间

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <utility>
#include <map>
#include <stack>
#include <set>
#include <vector>
#include <queue>
#include <deque>
#include <sstream>
#define x first
#define y second
#define mp make_pair
#define pb push_back
#define LL long long
#define Pair pair<int,int>
#define LOWBIT(x) x & (-x)
using namespace std;

const int MOD=1e9+7;
const int INF=0x7ffffff;
const int magic=348;

struct node
{
int to,cap,open,close;
};
vector<node> v[548];

int n,e,s,t;
int dist[548];
priority_queue<Pair> q;

int calc(int time,node edge)
{
int res=0;
if (time%(edge.open+edge.close)>edge.open)
{
res+=(edge.open+edge.close)-time%(edge.open+edge.close);
res+=edge.cap;
return res;
}
if (edge.open-time%(edge.open+edge.close)>=edge.cap)
return edge.cap;
else
{
res+=edge.open-time%(edge.open+edge.close);
res+=edge.close;
res+=edge.cap;
return res;
}
}

void dijkstra()
{
int i,x,y,dd;node edge;
for (i=1;i<=n;i++) dist[i]=INF;
dist[s]=0;q.push(mp(0,s));
while (!q.empty())
{
x=q.top().y;dd=-q.top().x;q.pop();
if (dd>dist[x]) continue;
//cout<<x<<endl;
for (i=0;i<v[x].size();i++)
{
edge=v[x][i];y=edge.to;
int tt=calc(dist[x],edge);
//cout<<tt<<endl;
//cout<<dist[y]<<' '<<dist[x]+tt<<endl;
if (dist[y]>dist[x]+tt)
{
dist[y]=dist[x]+tt;
q.push(mp(-dist[y],y));
}
}
}
}

int main ()
{
int i,ca=0,start;node ins;
while (scanf("%d%d%d%d",&n,&e,&s,&t)!=EOF)
{
for (i=1;i<=n;i++) v[i].clear();
for (i=1;i<=e;i++)
{
scanf("%d",&start);
scanf("%d%d%d%d",&ins.to,&ins.open,&ins.close,&ins.cap);
if (ins.cap>ins.open) continue;
v[start].pb(ins);
}
dijkstra();
//for (i=1;i<=n;i++) cout<<dist[i]<<' ';
//cout<<endl;
printf("Case %d: %d\n",++ca,dist[t]);
}
return 0;
}