SGU 103. Traffic Lights 带限制最短路

每个点有2中颜色 只有一条路上的两个点颜色一样才能通过这条路 最短路加上等待的时间处理 处理的是参考别人的 唉还是太弱了

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <algorithm>using namespace std;int s, e;int n, m;int a[333];int b[333];int c[333];int first[333], cnt;struct edge{int v, w, next;}G[30000];int d[333];int p[333];bool vis[333];struct HeapNode{int v, d;HeapNode(){}HeapNode(int v, int d): v(v), d(d){}bool operator < (const HeapNode& rhs) const{return d > rhs.d;}};void AddEdge(int u, int v, int w){G[cnt].v = v;G[cnt].w = w;G[cnt].next = first[u];first[u] = cnt++;G[cnt].v = u;G[cnt].w = w;G[cnt].next = first[v];first[v] = cnt++;}int wait(int t, int x, int y){int ans = 0;int t1 = (a[x] + t) % c[x];int t2 = (a[y] + t) % c[y];int tmp;for(int i = 0; i < 4; i++){if((t1 < b[x]) == (t2 < b[y]))return ans;if(t1 < b[x])tmp = b[x] - t1;elsetmp = c[x] - t1;if(t2 < b[y])tmp = min(tmp, b[y] - t2);elsetmp = min(tmp, c[y] - t2);t1 = (t1 + tmp) % c[x];t2 = (t2 + tmp) % c[y];ans += tmp;}return -1;}void print(int x){if(p[x] == -1){printf("%d", x);return;}print(p[x]);printf(" %d", x);}void Dijkstra(){for(int i = 1; i <= n; i++){d[i] = 999999999;p[i] = -1;vis[i] = false;}d[s] = 0;priority_queue <HeapNode> Q;Q.push(HeapNode(s, 0));while(!Q.empty()){HeapNode x = Q.top(); Q.pop();int u = x.v;if(vis[u])continue;vis[u] = 1;for(int i = first[u]; i != -1; i = G[i].next){int v = G[i].v;int x = d[u];int t = wait(x, u, v);if(t != -1 && d[v] > d[u] + G[i].w + t){d[v] = d[u] + G[i].w + t;p[v] = u;Q.push(HeapNode(v, d[v]));}}}if(d[e] != 999999999){printf("%d\n", d[e]);print(e);puts("");}elseputs("0");}int main(){memset(first, -1, sizeof(first));cnt = 0;scanf("%d %d", &s, &e);scanf("%d %d", &n, &m);for(int i = 1; i <= n; i++){char s[10];int r, bl, pu;scanf("%s %d %d %d", s, &r, &bl, &pu);if(s[0] == 'B'){a[i] = bl-r;b[i] = bl;c[i] = bl+pu;}else{a[i] = bl+pu-r;b[i] = bl;c[i] = bl+pu;}}for(int i = 1; i <= m; i++){int u, v, w;scanf("%d %d %d", &u, &v, &w);AddEdge(u, v, w);}Dijkstra();return 0;}