POJ 3159 Candies——差分约束

题目中的关系可以抽象成为A->B的一条有向边，然后跑1~n的最短路即可（dijkstra）

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 3 * 1e5 + 10;
int n, m, tot, head[maxn], dis[maxn];
struct Node {
int u, dis;
Node(int uu, int dd) : u(uu), dis(dd) {}
bool operator < (const Node &node) const {
return dis > node.dis;
}
};
struct Edge {
int to, val, next;
}edge[maxn];
void addedge(int u, int v, int val) {
tot++;
edge[tot].to = v;
edge[tot].val = val;
edge[tot].next = head[u];
head[u] = tot;
}
int dijkstra(int s, int e) {
for (int i = 1; i <= n; i++) dis[i] = INF;
dis[s] = 0;
priority_queue<Node> q; q.push(Node(s, 0));
while (!q.empty()) {
Node node = q.top(); q.pop();
int u = node.u;
if (dis[u] < node.dis) continue;
for (int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].to, val = edge[i].val;
if (dis[v] > dis[u] + val) {
dis[v] = dis[u] + val;
q.push(Node(v, dis[v]));
}
}
}
return dis[e];
}
int main() {
scanf("%d %d", &n, &m);
memset(head, -1, sizeof(head));
while (m--) {
int u, v, val; scanf("%d %d %d", &u, &v, &val);
addedge(u, v, val);
}
printf("%d\n", dijkstra(1, n));
return 0;
}