HDU 4725(最短路之建图难点)

题意：n个点，每个点属于特定一层，相邻的两层之间距离固定是c，另有额外m条边，然后求1到n的最短路径，没有则输出-1。

思路：建图时出现了一点问题，SPFA和DJ都一直TLE，我原本的错误思路是：将每一层抽象出来，对应为n+1...n+n，然后读每个点所在层flo时，使i和第n+flo建边(双向边)，然后通过一个vis数组标记第i层是否出现过，然后再对同时出现的相邻两层进行建边，本感觉思路恰好，肯定能过，但这样建图会存在一个问题，会导致同一层的所有点之间的距离均为1！！所以同一层的点如果想通过层到达，需要先到别的层，再回来此层。

所以建图时可直接建一个从第flo层到点i的单向边表示可从flo层回到i，加两条i到邻层的单向边(邻层存在则建，不存在不建)，表示i通过层的关系到达邻层的点，同时也要建层与层之间的边，原因：当上一层的点到达该层，此时可以走0距离走向该层的点，也可走c距离再向下一层走。

完整思路：层与层建边，点与点建边，层与在该层上的点建边 （边长为0），点与相邻层(存在的相邻层)建边 （边长为c）。避免了拆点的思路（拆点方法可看kuangbin的博客园）。

代码1

#include <algorithm>#include <iostream>#include <string.h>#include <cstdio>#include <queue>using namespace std;const int inf = 0x3f3f3f3f;const int maxn = 1e5+5;struct node{int v, w, next;}edge[maxn*8];int n, m, c, no;int head[maxn*2], book[maxn*2], dis[maxn*2], flr[maxn], vis[maxn];queue<int> q;void init(){no = 0;memset(book, 0, sizeof book);memset(dis, 0x3f, sizeof dis);memset(head, -1, sizeof head);memset(vis, 0, sizeof vis);}void add(int u, int v, int w){edge[no].v = v;edge[no].w = w;edge[no].next = head[u];head[u] = no++;}int SPFA(int s, int t){while(!q.empty()) q.pop();dis[s] = 0;q.push(s);book[s] = 1;while(!q.empty()){int tp = q.front(); q.pop();book[tp] = 0;int k = head[tp];while(k != -1){if(dis[edge[k].v] > dis[tp] + edge[k].w){dis[edge[k].v] = dis[tp] + edge[k].w;if(book[edge[k].v] == 0){book[edge[k].v] = 1;q.push(edge[k].v);}}k = edge[k].next;}}if(dis[t] == inf) return -1;return dis[t];}int main(){int t, flo, u, v, w;scanf("%d", &t);for(int l = 1; l <= t; ++l){scanf("%d %d %d", &n, &m, &c);init();for(int i = 1; i <= n; ++i){scanf("%d", &flo);flr[i] = flo;vis[flo] = 1;}for(int i = 2; i <= n; ++i){if(vis[i] && vis[i-1]) {add(n+i, n+i-1, c);add(n+i-1, n+i, c);}}for(int i = 1; i <= n; ++i){add(n+flr[i], i, 0);if(vis[flr[i]-1]) add(i, n+flr[i]-1, c);if(vis[flr[i]+1]) add(i, n+flr[i]+1, c);}for(int i = 1; i <= m; ++i){scanf("%d %d %d", &u, &v, &w);add(u, v, w);add(v, u, w);}printf("Case #%d: %d\n", l, SPFA(1, n));}return 0;}

代码2

#include <algorithm>#include <iostream>#include <string.h>#include <cstdio>#include <queue>using namespace std;const int inf = 0x3f3f3f3f;const int maxn = 1e5+5;struct node{int v, w, next;bool operator<(const node k) const{return w > k.w;}}edge[maxn*8];int n, m, c, no;int head[maxn*2], book[maxn*2], dis[maxn*2], flr[maxn], vis[maxn];priority_queue<node> q;void init(){no = 0;memset(book, 0, sizeof book);memset(dis, 0x3f, sizeof dis);memset(head, -1, sizeof head);memset(vis, 0, sizeof vis);}void add(int u, int v, int w){edge[no].v = v;edge[no].w = w;edge[no].next = head[u];head[u] = no++;}int DJ(int s, int t){while(!q.empty()) q.pop();dis[s] = 0;q.push((node){s, 0, -1});while(!q.empty()){node tp = q.top(); q.pop();if(book[tp.v]) continue;book[tp.v] = 1;int k = head[tp.v];while(k != -1){if(dis[edge[k].v] > dis[tp.v] + edge[k].w){dis[edge[k].v] = dis[tp.v] + edge[k].w;q.push((node){edge[k].v, dis[edge[k].v], -1});}k = edge[k].next;}}if(dis[t] == inf) return -1;return dis[t];}int main(){int t, flo, u, v, w;scanf("%d", &t);for(int l = 1; l <= t; ++l){scanf("%d %d %d", &n, &m, &c);init();for(int i = 1; i <= n; ++i){scanf("%d", &flo);flr[i] = flo;vis[flo] = 1;}for(int i = 2; i <= n; ++i){if(vis[i] && vis[i-1]) {add(n+i, n+i-1, c);add(n+i-1, n+i, c);}}for(int i = 1; i <= n; ++i){add(n+flr[i], i, 0);            if(flr[i] > 1) add(i, n+flr[i]-1, c);            if(flr[i] < n) add(i, n+flr[i]+1, c);}for(int i = 1; i <= m; ++i){scanf("%d %d %d", &u, &v, &w);add(u, v, w);add(v, u, w);}printf("Case #%d: %d\n", l, DJ(1, n));}return 0;}

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