Flight
Recently, Shua Shua had a big quarrel with his GF. He is so upset that he decides to take a trip to some other city to avoid meeting her. He will travel only by air and he can go to any city if there exists a flight and it can help him reduce the total cost to the destination. There‘s a problem here: Shua Shua has a special credit card which can reduce half the price of a ticket ( i.e. 100 becomes 50, 99 becomes 49. The original and reduced price are both integers. ). But he can only use it once. He has no idea which flight he should choose to use the card to make the total cost least. Can you help him?
InputThere are no more than 10 test cases. Subsequent test cases are separated by a blank line.
The first line of each test case contains two integers N and M ( 2 <= N <= 100,000
0 <= M <= 500,000 ), representing the number of cities and flights. Each of the following M lines contains "X Y D" representing a flight from city X to city Y with ticket price D ( 1 <= D <= 100,000 ). Notice that not all of the cities will appear in the list! The last line contains "S E" representing the start and end city. X, Y, S, E are all strings consisting of at most 10 alphanumeric characters.
OutputOne line for each test case the least money Shua Shua have to pay. If it‘s impossible for him to finish the trip, just output -1.Sample Input
4 4 Harbin Beijing 500 Harbin Shanghai 1000 Beijing Chengdu 600 Shanghai Chengdu 400 Harbin Chengdu 4 0 Harbin Chengdu
Sample Output
800 -1
Hint
In the first sample, Shua Shua should use the card on the flight from Beijing to Chengdu, making the route Harbin->Beijing->Chengdu have the least total cost 800. In the second sample, there‘s no way for him to get to Chengdu from Harbin, so -1 is needed. 题意：ShuaShua要从一个城市到另一个城市，给出每两城市之间的花费（有向），ShuaShua可以有一次半价的机会，求最小花费。思路：最短路径问题。开始想到求出最短路后选择其中最大的边/2，但其实错误，很容易举出反例：路径1：1 1 100 路径2：30 30 30 开始时102 90选择花费少的路径2，减半价后52 75路径1花费反而相对更少。因此我们可以换一种思路，以起点和终点为单源分别求出到各点的最短路，然后枚举每一条边作为中间边，dis[u]+w(u,v)/2+diss[v]的最小值为解。
#include<stdio.h>
#include<string.h>
#include<vector>
#include<deque>
#include<map>
#include<string>
#define MAX 100005
#define MAXX 500005
#define INF 10000000000000000
using namespace std;
struct Node{
int v,w;
}node;
vector <Node> edge[MAX],redge[MAX];
map<string,int> mp;
long long dis[MAX],diss[MAX],b[MAX],u[MAXX],v[MAXX],w[MAXX];
char s1[MAX],s2[MAX];
int n;
void spfa1(int k)
{
int i;
deque<int> q;
for(i=1;i<=n;i++){
dis[i]=INF;
}
memset(b,0,sizeof(b));
b[k]=1;
dis[k]=0;
q.push_back(k);
while(q.size()){
int u=q.front();
for(i=0;i<edge[u].size();i++){
int v=edge[u][i].v;
long long w=edge[u][i].w;
if(dis[v]>dis[u]+w){
dis[v]=dis[u]+w;
if(b[v]==0){
b[v]=1;
if(dis[v]>dis[u]) q.push_back(v);
else q.push_front(v);
}
}
}
b[u]=0;
q.pop_front();
}
}
void spfa2(int k)
{
int i;
deque<int> q;
for(i=1;i<=n;i++){
diss[i]=INF;
}
memset(b,0,sizeof(b));
b[k]=1;
diss[k]=0;
q.push_back(k);
while(q.size()){
int u=q.front();
for(i=0;i<redge[u].size();i++){
int v=redge[u][i].v;
long long w=redge[u][i].w;
if(diss[v]>diss[u]+w){
diss[v]=diss[u]+w;
if(b[v]==0){
b[v]=1;
if(diss[v]>diss[u]) q.push_back(v);
else q.push_front(v);
}
}
}
b[u]=0;
q.pop_front();
}
}
int main()
{
int m,bg,ed,t,i,j;
while(~scanf("%d%d",&n,&m)){
t=0;
for(i=1;i<=n;i++){
edge[i].clear();
redge[i].clear();
mp.clear();
}
memset(u,0,sizeof(u));
memset(v,0,sizeof(v));
memset(w,0,sizeof(w));
for(i=1;i<=m;i++){
scanf(" %s%s%lld",s1,s2,&w[i]);
if(!mp[s1]) mp[s1]=++t;
if(!mp[s2]) mp[s2]=++t;
u[i]=mp[s1];
v[i]=mp[s2];
node.v=mp[s2];
node.w=w[i];
edge[mp[s1]].push_back(node);
node.v=mp[s1];
redge[mp[s2]].push_back(node);
}
scanf(" %s%s",s1,s2);
if(!mp[s1]) mp[s1]=++t;
if(!mp[s2]) mp[s2]=++t;
bg=mp[s1],ed=mp[s2];
spfa1(bg);
spfa2(ed);
long long min=INF;
for(i=1;i<=m;i++){
if(dis[u[i]]==INF||dis[v[i]]==INF) continue;
if(dis[u[i]]+diss[v[i]]+w[i]/2<min) min=dis[u[i]]+diss[v[i]]+w[i]/2;
}
if(min==INF) printf("-1\n");
else printf("%lld\n",min);
}
return 0;
}