CF95C Volleyball
题意翻译
给出一个图,双向边,边上有权值代表路的距离,然后每个点上有两个值,t,c,t代表能从这个点最远沿边走t,且不能在半路下来,花费是c 现在告诉你起点终点,问最少的花费 点个数1000,边个数1000,边权1e9
By @partychicken
题目描述
Petya loves volleyball very much. One day he was running late for a volleyball match. Petya hasn't bought his own car yet, that's why he had to take a taxi. The city has n n n junctions, some of which are connected by two-way roads. The length of each road is defined by some positive integer number of meters; the roads can have different lengths.
Initially each junction has exactly one taxi standing there. The taxi driver from the i i i -th junction agrees to drive Petya (perhaps through several intermediate junctions) to some other junction if the travel distance is not more than ti t_{i} ti meters. Also, the cost of the ride doesn't depend on the distance and is equal to ci c_{i} ci bourles. Taxis can't stop in the middle of a road. Each taxi can be used no more than once. Petya can catch taxi only in the junction, where it stands initially.
At the moment Petya is located on the junction x x x and the volleyball stadium is on the junction y y y . Determine the minimum amount of money Petya will need to drive to the stadium.
输入输出格式
输入格式:
The first line contains two integers n n n and m m m ( 1<=n<=1000,0<=m<=1000) 1<=n<=1000,0<=m<=1000) 1<=n<=1000,0<=m<=1000) . They are the number of junctions and roads in the city correspondingly. The junctions are numbered from 1 1 1 to n n n , inclusive. The next line contains two integers x x x and y y y ( 1<=x,y<=n 1<=x,y<=n 1<=x,y<=n ). They are the numbers of the initial and final junctions correspondingly. Next m m m lines contain the roads' description. Each road is described by a group of three integers ui u_{i} ui , vi v_{i} vi , wi w_{i} wi ( 1<=ui,vi<=n,1<=wi<=109 1<=u_{i},v_{i}<=n,1<=w_{i}<=10^{9} 1<=ui,vi<=n,1<=wi<=109 ) — they are the numbers of the junctions connected by the road and the length of the road, correspondingly. The next n n n lines contain n n n pairs of integers ti t_{i} ti and ci c_{i} ci ( 1<=ti,ci<=109 1<=t_{i},c_{i}<=10^{9} 1<=ti,ci<=109 ), which describe the taxi driver that waits at the i i i
-th junction — the maximum distance he can drive and the drive's cost.
The road can't connect the junction with itself, but between a pair of
junctions there can be more than one road. All consecutive numbers in
each line are separated by exactly one space character.
输出格式:
If taxis can't drive Petya to the destination point, print "-1" (without the quotes). Otherwise, print the drive's minimum cost.
Please do not use the %lld specificator to read or write 64-bit
integers in С++. It is preferred to use cin, cout streams or the %I64d
specificator.
输入输出样例
4 4
1 3
1 2 3
1 4 1
2 4 1
2 3 5
2 7
7 2
1 2
7 7
9
说明
An optimal way — ride from the junction 1 to 2 (via junction 4), then from 2 to 3. It costs 7+2=9 bourles.
Solution:
本题水。
点数很小,先从每个点暴力最短路处理出该点的t范围内能到的点,并且建一张新图,然后只要在新图上再跑一遍最短路就好了。
代码:
/*Code by 520 -- 8.21*/
#include<bits/stdc++.h>
#define il inline
#define ll long long
#define RE register
#define For(i,a,b) for(RE int (i)=(a);(i)<=(b);(i)++)
#define Bor(i,a,b) for(RE int (i)=(b);(i)>=(a);(i)--)
using namespace std;
const int N=;
int n,m,s,t,a[N],b[N];
int to[N],net[N],w[N],h[N],cnt;
int To[N],Net[N],W[N],H[N],Cnt;
ll dis[N];
bool vis[N];
struct node{
int u;
ll d;
bool operator<(const node &a)const{return d>a.d;}
}; int gi(){
int a=;char x=getchar();
while(x<''||x>'')x=getchar();
while(x>=''&&x<='')a=(a<<)+(a<<)+(x^),x=getchar();
return a;
} il void add(int u,int v,int c){to[++cnt]=v,net[cnt]=h[u],w[cnt]=c,h[u]=cnt;} il void Add(int u,int v,int c){To[++cnt]=v,Net[cnt]=H[u],W[cnt]=c,H[u]=cnt;} queue<int>q;
il void spfa(int x){
For(i,,n) dis[i]=;
q.push(x),dis[x]=;
while(!q.empty()){
int u=q.front();q.pop();vis[u]=;
for(RE int i=h[u];i;i=net[i])
if(dis[to[i]]>dis[u]+w[i]){
dis[to[i]]=dis[u]+w[i];
if(!vis[to[i]])q.push(to[i]),vis[to[i]]=;
}
}
For(i,,n) if(i!=x&&dis[i]<=a[x]) Add(x,i,b[x]);
} priority_queue<node>Q;
il void dij(){
For(i,,n) dis[i]=;
dis[s]=,Q.push(node({s,}));
while(!Q.empty()){
node x=Q.top();Q.pop();
if(!vis[x.u]){
vis[x.u]=;
for(RE int i=H[x.u];i;i=Net[i])
if(dis[To[i]]>dis[x.u]+W[i]){
dis[To[i]]=dis[x.u]+W[i];
if(!vis[To[i]]) Q.push(node({To[i],dis[To[i]]}));
}
}
}
} il void init(){
n=gi(),m=gi(),s=gi(),t=gi();
int u,v,c;
For(i,,m) u=gi(),v=gi(),c=gi(),add(u,v,c),add(v,u,c);
For(i,,n) a[i]=gi(),b[i]=gi();
cnt=;
For(i,,n) spfa(i);
dij();
cout<<(dis[t]==?-:dis[t]);
} int main(){
init();
return ;
}
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