题目链接:hdu 4725 The Shortest Path in Nya Graph
题目大意:n个点,m条边,以及相邻层之间移动的代价c,给出每个点所在的层数,以及m条边,每条边有u,v,val,表示从节点u到v(无向),并且移动的代价val,问说从1到n的代价最小是多少。
解题思路:dijkstra算法,主要是建图,每一层有一个汇点,汇点连接所有处于当前层的点,代价为0,相邻层之间如果两层都有点的话,连接,代价为c,然后每个点连接上下层的汇点,代价c(这里我一开始写将当前点连接到当前层的汇点,结果WA了,原因是因为如果存在两个点位于同一层,那么它们到大层汇点的代价都为0,也就是说明两点之间移动的代价为0,所以和原先的图有出入)。剩下的就是dijkstra算了。
#include <cstdio> #include <cstring> #include <cmath> #include <vector> #include <queue> using namespace std; typedef pair<int,int> pii; const int N = 1e5; const int INF = 0x3f3f3f3f; int n, m, c, a[N*3+10], vis[N*3+10]; int nindex, g[N*3+10]; struct state { int v; int val; int next; }edge[N*20]; inline void addEdge (int u, int v, int val) { edge[nindex].v = v; edge[nindex].val = val; edge[nindex].next = g[u]; g[u] = nindex++; } void init () { nindex = 0; memset(g, -1, sizeof(g)); memset(vis, 0, sizeof(vis)); scanf("%d%d%d", &n, &m, &c); int a, b, d; for (int i = 1; i <= n; i++) { scanf("%d", &d); vis[d] = 1; // addEdge(i, d+N, 0); addEdge(d+N, i, 0); if (d > 1) addEdge(i, d + N - 1, c); if (d < n) addEdge(i, d + N + 1, c); } for (int i = 0; i < m; i++) { scanf("%d%d%d", &a, &b, &d); addEdge(a, b, d); addEdge(b, a, d); } for (int i = 2; i <= n; i++) { if (vis[i] && vis[i-1]) { addEdge(i+N, i-1+N, c); addEdge(i-1+N, i+N, c); } } } int dijkstra() { memset(a, INF, sizeof(a)); priority_queue< pii, vector<pii>, greater<pii> > que; a[1] = 0; que.push(make_pair(a[1], 1)); while ( !que.empty()) { pii cur = que.top(); que.pop(); int x = cur.second; for (int j = g[x]; j != -1; j = edge[j].next) { int u = edge[j].v; if (a[u] > a[x] + edge[j].val) { a[u] = a[x] + edge[j].val; que.push(make_pair(a[u], u)); } } } if (a[n] == INF) return -1; else return a[n]; } int main () { int cas; scanf("%d", &cas); for (int i = 1; i <= cas; i++) { init(); printf("Case #%d: %d\n", i, dijkstra()); } return 0; }