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;
}

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