Age of Moyu
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 3125 Accepted Submission(s): 981
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
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其实当时没怎么认真思考这道题要怎么写
补题时候看的题解
有个题解说用set 其实没怎么搞清楚要怎么写 但是这种方法也可以实现
用优先队列优化dijkstra 队列维护的是边而不是点
根据当前边的编号决定cost是否增加
因为是优先队列所以可以保证最先到达n的线路是cost最小的
好像比赛的时候他们写T了好多次....不知道他们是怎么写的
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<queue>
#define inf 0x3f3f3f3f
using namespace std;
int n, m;
const int maxn = 100005;
const int maxm = 200005;
int head[maxn], cnt;
struct edge{
int v, f, w;
int nxt;
bool operator < (edge a) const{
return w > a.w;
}
}e[maxm * 2];
void addedge(int u, int v, int f)
{
e[cnt].v = v;
e[cnt].f = f;
e[cnt].nxt = head[u];
head[u] = cnt;
cnt++;
}
int dis[maxn], vis[maxn];
void init()
{
cnt = 0;
memset(vis, 0, sizeof(vis));
memset(dis, inf, sizeof(dis));
memset(head, -1, sizeof(head));
}
void dijkstra()
{
dis[1] = 0;
priority_queue <edge> q;
while(!q.empty()){
q.pop();
}
edge now;
now.v = 1;
now.f = 0;
now.w = 0;
q.push(now);
while(!q.empty()){
now = q.top();
q.pop();
int u = now.v;
if(vis[u]){
continue;
}
vis[u] = true;
dis[u] = now.w;
for(int i = head[u]; ~i; i = e[i].nxt){
int v = e[i].v;
if(!vis[v]){
edge nex;
nex.v = v;
nex.f = e[i].f;
nex.w = dis[u] + (e[i].f != now.f);
q.push(nex);
}
}
}
}
int main()
{
while(scanf("%d%d", &n, &m) != EOF){
init();
for(int i = 0; i < m; i++){
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
addedge(a, b, c);
addedge(b, a, c);
}
dijkstra();
if(dis[n] == inf){
printf("-1
");
}
else{
printf("%d
", dis[n]);
}
}
return 0;
}