(图论)追债之旅
作者:互联网
\(最短路 + dp\)
思路
\(这里用了拆点,即每个点并不只是位置,而是位置+到达的天数\)
\(dis[i][j]:表示第i天到达j的最小花费\)
\(dist[day][to]=min(dist[day][to],dist[day-1][now]+w[i]+cosy[day])\)
\(由于拆点,我们需要重载一个Node\)
代码
#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);
inline int lowbit(int x) { return x & (-x); }
#define ll long long
#define ull unsigned long long
#define pb push_back
#define PII pair<int, int>
#define VIT vector<int>
#define x first
#define y second
#define inf 0x3f3f3f3f
const int N = 1010, M = 1e4 + 10;
int h[N], ne[M], e[M], w[M], idx;
bool st[15][N];
int dist[15][N];
int a[15];
int n, m, k;
void add(int a, int b, int c) {
e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}
struct Node {
int day, pos, dis;
Node(int _day, int _pos, int _dis) : day(_day), pos(_pos), dis(_dis) {}
bool operator < (const Node & t) const {
return dis > t.dis;
}
};
void dijkstra() {
memset(dist, 0x3f, sizeof dist);
memset(st, 0, sizeof st);
dist[0][1] = 0;
priority_queue<Node> q;
q.push(Node(0, 1, 0));
while (q.size()) {
auto t = q.top();
q.pop();
int day = t.day, pos = t.pos, dis = t.dis;
if (st[day][pos]) continue;
st[day][pos] = true;
for (int i = h[pos]; ~i; i = ne[i]) {
if (day + 1 > k) continue;
int j = e[i];
if (dist[day + 1][j] > dis + a[day + 1] + w[i]) {
dist[day + 1][j] = dis + a[day + 1] + w[i];
q.push(Node(day + 1, j, dist[day + 1][j]));
}
}
}
}
int main() {
IO;
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
memset(h, -1, sizeof h);
cin >> n >> m >> k;
while (m--) {
int a, b, c;
cin >> a >> b >> c;
add(a, b, c), add(b, a, c);
}
for (int i = 1; i <= k; ++i) cin >> a[i];
dijkstra();
int ans = inf;
for (int i = 1; i <= k; ++i) ans = min(ans, dist[i][n]);
if (ans == inf) cout << "-1\n";
else cout << ans << '\n';
}
标签:图论,dist,之旅,int,pos,追债,define,day,dis 来源: https://www.cnblogs.com/phr2000/p/14464382.html