CSU 1808：地铁（Dijkstra）

http://acm.csu.edu.cn/csuoj/problemset/problem?pid=1808

题意：……

思路：和之前的天梯赛的一题一样，但是简单点。

没办法直接用点去算。把边看成点去做，规定dis[i]为走完第i条边之后即达到edge[i].v这个点的时候需要的花费。

点数为2*m。如果用普通的Dijkstra和SPFA会超时，所以用优先队列优化的Dijkstra。

#include <bits/stdc++.h>
using namespace std;
#define N 100010
typedef long long LL;
const LL INF = 1000000000000000000LL;
struct Edge {
    int u, v, nxt, w, c;
} edge[N*2];
struct Node {
    LL d; int id;
    bool operator < (const Node &rhs) const {
        return d > rhs.d;
    }
};
int head[N], tot, n, m;
bool vis[N*2];
LL dis[N*2];

void Add(int u, int v, int w, int id) {
    edge[tot] = (Edge) { u, v, head[u], w, id}; head[u] = tot++;
    edge[tot] = (Edge) { v, u, head[v], w, id}; head[v] = tot++;
}

LL Dijkstra() {
    for(int i = 0; i < tot; i++) dis[i] = INF;
    memset(vis, 0, sizeof(vis));
    LL ans = INF;
    priority_queue<Node> que;
    while(!que.empty()) que.pop();

    for(int i = head[1]; ~i; i = edge[i].nxt)
        que.push((Node) {edge[i].w, i}), dis[i] = edge[i].w;
    while(!que.empty()) {
        Node now = que.top(); que.pop();
        int pree = now.id; LL pred = now.d;
        if(vis[pree]) continue; vis[pree] = 1;
        int u = edge[pree].v;
        if(u == n && ans > pred) ans = pred;
        for(int i = head[u]; ~i; i = edge[i].nxt) {
            int nowe = i;
            LL nowd = dis[pree] + edge[nowe].w + abs(edge[nowe].c - edge[pree].c);
            if(nowd < dis[nowe] && !vis[nowe]) {
                dis[nowe] = nowd;
                que.push((Node) { nowd, nowe });
            }
        }
    }
    return ans;
}

int main() {
    while(~scanf("%d%d", &n, &m)) {
        memset(head, -1, sizeof(head)); tot = 0;
        for(int i = 1; i <= m; i++) {
            int u, v, c, w;
            scanf("%d%d%d%d", &u, &v, &c, &w);
            Add(u, v, w, c);
        }
        printf("%lld
", Dijkstra());
    }
    return 0;
}
/*
3 3
1 2 1 1
2 3 2 1
1 3 1 1
3 3
1 2 1 1
2 3 2 1
1 3 1 10
3 2
1 2 1 1
2 3 1 1
*/