HDU - 4725 topic Link
Subject to the effect
This is a hierarchical diagram.
- Point between layers can reach each other, but if there is no edge, it can not be connected directly to the point in the same layer.
- There are some points between the connected side, you can communicate directly without going through the floor, of course, also be faster if the floor by then.
Difficulties and handling
m side edges to establish simple, the key point is how to handle the relationship between good layers.
We additionally introduced n points serve as floors, number n + 1 ~ n + n
we establish direct current floor edge point, attention must be the same floor where the sides Unidirectional
Below, we assume that the first layer 2 point, 3 point, the second layer, and there is no edge between each point can reach each other, the cost spent on each floor
of from 1 -> 2 is only one way, go from point 1 of the first layer, the second layer 3 point, and then went to the point of the first layer 2 second layer 3 by a dot, is cost 2cost
If we establish a point on the floor and the floor side two-way, point 1-- spending is 0> 2 a.
Therefore, we must establish a unified side between points on the floor and then the floor, from the floor to point, point to the floor or from
//Powered by CK 2020:04:08
#include<bits/stdc++.h>
using namespace std;
typedef pair<int, int> PII;
const int INF = 0x3f3f3f3f;
const int N = 1e6 + 10;
int head[N], to[N], value[N], nex[N], cnt;
int visit[N], dis[N], n, m, cost;
struct cmp {
bool operator () (const PII & a, const PII & b) const {
return a.second > b.second;
}
};
void add(int x, int y, int w) {
to[cnt] = y;
value[cnt] = w;
nex[cnt] = head[x];
head[x] = cnt++;
}
int Dijkstra() {
memset(dis, 0x3f, sizeof dis);
memset(visit, 0, sizeof visit);
priority_queue<PII, vector<PII>, cmp> q;
q.push(make_pair(1, 0));
dis[1] = 0;
while(!q.empty()) {
int temp = q.top().first;
q.pop();
if(visit[temp]) continue;
visit[temp] = 1;
for(int i = head[temp]; i; i = nex[i]) {
if(dis[to[i]] > dis[temp] + value[i]) {
dis[to[i]] = dis[temp] + value[i];
q.push(make_pair(to[i], dis[to[i]]));
}
}
}
return dis[n] == INF ? -1 : dis[n];
}
int main() {
// freopen("in.txt", "r", stdin);
int t, x, y, w;
scanf("%d", &t);
for(int cas = 1; cas <= t; cas++) {
printf("Case #%d: ", cas);
scanf("%d %d %d", &n, &m, &cost);
memset(head, 0, sizeof head);
cnt = 1;
for(int i = 1; i<= n; i++) {
scanf("%d", &x);
add(i, x + n, 0);//点到楼层
// add(x + n, i, 0);//楼层到点————二选一
if(x - 1 >= 1) add(x - 1 + n, i, cost), add(i, x - 1 + n, cost);
if(x + 1 <= n) add(x + 1 + n, i, cost), add(i, x + 1 + n, cost);
}
for(int i = 0; i < m; i++) {
scanf("%d %d %d", &x, &y, &w);
add(x, y, w);
add(y, x, w);
}
printf("%d\n", Dijkstra());
}
return 0;
}