ZOJ1232 Adventure of Super Mario spfa上的dp

很早之前听说有一种dp是在图上的dp，然后是在跑SPFA的时候进行dp，所以特地找了一题关于在SPFA的时候dp的。

题意：1~a是村庄 a+1~a+b是城堡，存在m条无向边。求由a+b->1的最短路，但是你有很多飞鞋，每双飞鞋可以跑一个固定的距离l,但是跑的时候要是碰到了城堡就要停下来，而且也不能停在路中间。

思路和代码参考了下面的这个网址：http://blog.csdn.net/acm_cxlove/article/details/8679230

思路：正常来说我们就跑SPFA就可以了，每次从队列里取点更新，但因为有了跑鞋，所以每次取点的时候还有多一种转移途径，因此我们就需要知道每次用了跑鞋之后从当前的点能够到达哪些点，这个时候我们就需要两点之间的距离，因为两点之间不能经过城堡，所以在跑SPFA的时候我们对于那些城堡的点我们就不再压入的跑SPFA.然后当dis[u][v]<=l的时候我们就可以转移了。

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#pragma warning(disable:4996) #include<cstring> #include<cstdio> #include<iostream> #include<string> #include<algorithm> #include<vector> #include<queue> #define maxn 120 #define maxm 120*120 #define inf 0x3f3f3f3f using namespace std;   int a, b, m, l, k;   struct Edge {     int u, v, w;     Edge(){}     Edge(int ui, int vi, int wi) :u(ui), v(vi), w(wi){} }e[maxm]; int ecnt;   int dis[maxn][maxn]; int vis[maxn]; int first[maxn]; int nxt[maxm];   void add(int u, int v, int w) {     e[ecnt].u = u; e[ecnt].v = v; e[ecnt].w = w;     nxt[ecnt] = first[u];     first[u] = ecnt++; }   void spfa() {     memset(dis, 0x3f, sizeof(dis));     queue<int> que;     for (int loc = 1; loc <= a + b; loc++){         while (!que.empty()) que.pop();         memset(vis, 0, sizeof(vis));         que.push(loc); vis[loc] = 1;         dis[loc][loc] = 0;         while (!que.empty()){             int u = que.front(); que.pop(); vis[u] = 0;             for (int i = first[u]; i != -1; i = nxt[i]){                 int v = e[i].v, w = e[i].w;                 if (dis[loc][v] > dis[loc][u] + w){                     dis[loc][v] = dis[loc][u] + w;                     if (v <= a&&!vis[v]){                         que.push(v); vis[v] = 1;                     }                 }             }         }     } }   int dp[maxn][15];   int main() {     int T; cin >> T;     while (T--)     {         scanf("%d%d%d%d%d", &a, &b, &m, &l, &k);         memset(first, -1, sizeof(first)); ecnt = 0;         int ui, vi, wi;         for (int i = 0; i < m; i++){             scanf("%d%d%d", &ui, &vi, &wi);             add(ui, vi, wi);             add(vi, ui, wi);         }         spfa();         memset(dp, 0x3f, sizeof(dp));         memset(vis, 0, sizeof(vis));         queue<int> que;         que.push(a + b);         dp[a + b][k] = 0;         vis[a + b] = 1;         while (!que.empty()){             int u = que.front(); que.pop(); vis[u] = 0;             for (int sh = 0; sh <= k; sh++){                 for (int i = first[u]; i != -1; i = nxt[i]){                     int v = e[i].v, w = e[i].w;                     if (dp[v][sh] > dp[u][sh] + w){                         dp[v][sh] = dp[u][sh] + w;                         if (!vis[v]){                             que.push(v); vis[v] = 1;                         }                     }                 }                 if (!sh) continue;                 for (int v = 1; v <= a + b; v++){                     if (u != v&&dis[u][v] <= l){                         if (dp[v][sh - 1] > dp[u][sh]){                             dp[v][sh - 1] = dp[u][sh];                             if (!vis[v]){                                 que.push(v); vis[v] = 1;                             }                         }                     }                 }             }         }         int ans = inf;         for (int i = 0; i <= k; i++){             ans = min(ans, dp[1][i]);         }         printf("%d\n", ans);     }     return 0; }

 

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