SGU 103 Traffic Lights(最短路)

這題不錯，是一個最短路，但是中間有一個限制條件，就是等待時間

首先先看一下爲什麼仍然滿足最短路

因爲最短路肯定是每個結點求出最早到達的時間，那麼其實不管有沒等待，從隊頭取出去轉移的肯定是最早的時間，仍然滿足轉移

那麼就是等待時間如何去計算的問題

其實就先寫一個函數，獲得當前的顏色，和到下一個顏色的時間

然後如果顏色相同就不用等

如果顏色不同，到下一個顏色時間又不同，就返回其中小的時間

如果扔相同，就可以往後找4次（因爲最多3次就進入循環節了），直到有一個時間不同，否則就是無限交替，這條路永遠不能走

代碼：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

const int N = 305;
const int M = 30005;
const int INF = 0x3f3f3f3f;

int s, t;
int n, m;

struct Node {
    int tp, c, r, t;
    void read(int tp) {
        this->tp = tp;
        scanf("%d%d%d", &c, &r, &t);
    }
} node[N];

struct Edge {
    int u, v, w;
    Edge() {}
    Edge(int u, int v, int w) {
        this->u = u;
        this->v = v;
        this->w = w;
    }
} edge[M];

int head[N], nxt[M], en;

void add_edge(int u, int v, int w) {
    edge[en] = Edge(u, v, w);
    nxt[en] = head[u];
    head[u] = en++;
}

int d[N], vis[N], p[N];

struct State {
    int u, w;
    State() {}
    State(int u, int w) {
        this->u = u;
        this->w = w;
    }
    bool operator < (const State &c) const {
        return w > c.w;
    }
};

void print(int u) {
    if (u == s) {
        printf("%d", u);
        return;
    }
    print(p[u]);
    printf(" %d", u);
}

void get(int now, Node u, int &ca, int &ta) {
    int c = u.c, r = u.r, t = u.t, tp = u.tp;
    if (now < c) {
        ca = tp;
        ta = c - now;
        return;
    }
    now -= c; tp = !tp;
    now %= (r + t);
    if (tp == 0) {
        if (now < r) {
            ca = tp;
            ta = r - now;
            return;
        }
        now -= r; tp = !tp;
        ca = tp;
        ta = t - now;
        return;
    } else {
        if (now < t) {
            ca = tp;
            ta = t - now;
            return;
        }
        now -= t; tp = !tp;
        ca = tp;
        ta = r - now;
        return;
    }
}

int cal(int now, int u, int v) {
    int ca, ta, cb, tb;
    get(now, node[u], ca, ta);
    get(now, node[v], cb, tb);
    if (ca == cb) return 0;
    if (ta != tb)
        return min(ta, tb);
    else {
        int ans = 0;
        int flag = 1;
        for (int i = 0; i < 4; i++) {
            ca = !ca;
            cb = !cb;
            ans += ta;
            if (ca == 0) ta = node[u].r;
            else ta = node[u].t;
            if (cb == 0) tb = node[v].r;
            else tb = node[v].t;
            if (ta != tb) {
                flag = 0;
                break;
            }
        }
        if (flag) return INF;
        return ans + min(ta, tb);
    }
}

void solve() {
    for (int i = 1; i <= n; i++) d[i] = INF;
    memset(vis, 0, sizeof(vis));
    priority_queue<State> Q;
    Q.push(State(s, 0));
    d[s] = 0;
    while (!Q.empty()) {
        State x = Q.top();
        int u = x.u;
        if (u == t) break;
        Q.pop();
        if (vis[u]) continue;
        vis[u] = 1;
        for (int i = head[u]; i + 1; i = nxt[i]) {
            int v = edge[i].v;
            int w = edge[i].w;
            int tmp = cal(x.w, u, v) + x.w + w;
            if (d[v] > tmp) {
                d[v] = tmp;
                p[v] = u;
                Q.push(State(v, d[v]));
            }
        }
    }
    if (d[t] == INF) printf("0\n");
    else {
        printf("%d\n", d[t]);
        print(t);
        printf("\n");
    }
}

int main() {
    while (~scanf("%d%d", &s, &t)) {
        en = 0;
        memset(head, -1, sizeof(head));
        scanf("%d%d", &n, &m);
        char s[2];
        for (int i = 1; i <= n; i++) {
            scanf("%s", s);
            node[i].read(s[0] == 'P');
        }
        int u, v, w;
        while (m--) {
            scanf("%d%d%d", &u, &v, &w);
            add_edge(u, v, w);
            add_edge(v, u, w);
        }
        solve();
    }
    return 0;
}