Age of Moyu
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2539 Accepted Submission(s): 764
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 Source
2018 Multi-University Training Contest 7
题意：从1走向n，每条边的边权都为1，每条边都有一个类型，走连续类型的边花费为1，输出最小的花费
思路：以BFS为主体，每搜到一个点，在DFS出所有与这条边类型相同的边。实际上就是先搜1步到达的点，在搜2步到达的点，每搜一条边就标记一下边，把点加入队列
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 100005;
const int MAXM = 400005;
typedef long long LL;
struct node1
{
int v,c,next;
}edge[MAXM];
int head[MAXN],cnt;
int n,flag,ans;
bool vis[MAXM];
inline void addedge(int u,int v,int c)
{
edge[cnt].v = v;
edge[cnt].c = c;
edge[cnt].next = head[u];
head[u] = cnt++;
};
inline int read()
{
char ch = getchar(); int x = 0, f = 1;
while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
while('0' <= ch && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();}
return x * f;
}
struct node2
{
int u,step;
node2(){}
node2(int _u,int _step){u = _u,step = _step;};
};
queue<node2> q;
void DFS(int u,int c,int step)
{
for(int i = head[u]; i != -1; i = edge[i].next) {
if(!vis[i] && c == edge[i].c) {
if(edge[i].v == n) {
flag = 1;
ans = step;
return;
}
q.push(node2(edge[i].v,step));
vis[i] = 1;
vis[i ^ 1] = 1;
DFS(edge[i].v,c,step);
if(flag) return;
}
}
}
void BFS()
{
int v;
struct node2 now;
while(!q.empty()) q.pop();
q.push(node2(1,0));
while(!q.empty()) {
now = q.front();
q.pop();
for(int i = head[now.u]; i != -1; i = edge[i].next) {
if(!vis[i]) {
v = edge[i].v;
if(v == n) {
flag = 1;
ans = now.step + 1;
return;
}
q.push(node2(v,now.step + 1));
vis[i] = 1;
vis[i ^ 1] = 1;
DFS(v,edge[i].c,now.step + 1);
if(flag) return;
}
}
}
}
int main(void)
{
int m;
int u,v,c;
while(scanf("%d %d",&n,&m) != EOF) {
memset(head,-1,sizeof(head));
cnt = 0;
for(int i = 0; i < m; ++i) {
u = read(),v = read(),c = read();
addedge(u,v,c);
addedge(v,u,c);
}
memset(vis,0,sizeof(vis));
flag = 0;
BFS();
if(flag) printf("%d\n",ans);
else printf("-1\n");
}
return 0;
}