SGU 103 Traffic Lights(最短路)

这题不错，是一个最短路，但是中间有一个限制条件，就是等待时间

首先先看一下为什么仍然满足最短路

因为最短路肯定是每个结点求出最早到达的时间，那么其实不管有没等待，从队头取出去转移的肯定是最早的时间，仍然满足转移

那么就是等待时间如何去计算的问题

其实就先写一个函数，获得当前的颜色，和到下一个颜色的时间

然后如果颜色相同就不用等

如果颜色不同，到下一个颜色时间又不同，就返回其中小的时间

如果扔相同，就可以往后找4次（因为最多3次就进入循环节了），直到有一个时间不同，否则就是无限交替，这条路永远不能走

代码：

#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;}