Age of Moyu
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1531 Accepted Submission(s): 467
Problem Description
Mr.Quin love fishes so much and Mr.Quin’s city has a nautical system,consisiting of N ports and M shipping lines. The ports are numbered 1 to N. Each line is occupied by a Weitian. Each Weitian has an identification number.
The i-th (1≤i≤M) line connects port Ai and Bi (Ai≠Bi) bidirectionally, and occupied by Ci Weitian (At most one line between two ports).
When Mr.Quin only uses lines that are occupied by the same Weitian, the cost is 1 XiangXiangJi. Whenever Mr.Quin changes to a line that is occupied by a different Weitian from the current line, Mr.Quin is charged an additional cost of 1 XiangXiangJi. In a case where Mr.Quin changed from some Weitian A's line to another Weitian's line changes to Weitian A's line again, the additional cost is incurred again.
Mr.Quin is now at port 1 and wants to travel to port N where live many fishes. Find the minimum required XiangXiangJi (If Mr.Quin can’t travel to port N, print −1instead)
Input
There might be multiple test cases, no more than 20. You need to read till the end of input.
For each test case,In the first line, two integers N (2≤N≤100000) and M (0≤M≤200000), representing the number of ports and shipping lines in the city.
In the following m lines, each contain three integers, the first and second representing two ends Ai and Bi of a shipping line (1≤Ai,Bi≤N) and the third representing the identification number Ci (1≤Ci≤1000000) of Weitian who occupies this shipping line.
Output
For each test case output the minimum required cost. If Mr.Quin can’t travel to port N, output −1 instead.
Sample Input
3 3 1 2 1 1 3 2 2 3 1 2 0 3 2 1 2 1 2 3 2
Sample Output
1 -1 2
题意：给你一个图，每条边有一种颜色。你每走一种颜色消耗1分钟，然后沿该颜色的边继续走就不消耗时间，走其他颜色的边就会再消耗1分钟。求从1点走到n点至少需要几分钟，如果走不到n点输出-1。
思路：就是最短路的变形。虽然这道题好像标程错了还是原题（但是出一道题毕竟不容易）。。。有多种解法，这里我只说我比赛的时候过了的解法。
由于保证全局最优，我们每次只走一步，然后暴力dfs遍历这一步能走到的所有点（这样一定保证1点到每一个点都是最短的）。由于所有边的点都只走一次，保证了时间复杂度不会超过O(n+m)自己xjb乱算的，为了节省时间到达n直接返回。
代码：
#pragma comment(linker,"/STACK:10240000,10240000")
#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=400010;
struct node
{
int id,u,v,cost;
node(int _id=0,int _u=0,int _v=0,int _cost=0):id(_id),u(_u),v(_v),cost(_cost){}
};
queue<int>q;
vector<node>e[maxn];
int n,m,k;
int dis[maxn],vis[maxn];
void addedge(int id,int u,int v,int w)
{
e[u].push_back(node(id,u,v,w));
}
void dfs(int x,int id,int dep,int fa)
{
if(!vis[x]){q.push(x);dis[x]=dep;vis[x]=1;}
if(dis[n]!=inf)return;
for(int i=0;i<e[x].size();i++)
{
node &k=e[x][i];
if(k.v==fa||k.id) continue;
if(k.cost==id)
{
k.id=1;
dfs(k.v,id,dep,x);
if(dis[n]!=inf)return;
}
}
}
int main()
{
int T,cas=1;
while(scanf("%d %d",&n,&m)!=EOF){
for(int i=0;i<=n;i++)
{dis[i]=inf;e[i].clear();vis[i]=0;}
while(!q.empty())q.pop();
int u,v,w;
for(int i=1;i<=m;i++)
{
scanf("%d%d%d",&u,&v,&w);
if(u==v) continue;
addedge(0,u,v,w);
addedge(0,v,u,w);
}
dis[1]=0;q.push(1);vis[1]=1;
while(!q.empty())
{
int k=q.front();q.pop();
for(int i=0;i<e[k].size();i++)
{
node &j=e[k][i];
if(j.id) continue;
j.id=1;
dfs(j.v,j.cost,dis[k]+1,-1);
if(dis[n]!=inf)break;
}
if(dis[n]!=inf) break;
}
if(dis[n]==inf) puts("-1");
else printf("%d\n",dis[n]);
}
return 0;
}