USACO Sweet Butter 解题报告

刚开始用Floyd Warshall算法求每对节点的最短路径，时间复杂度为v^3，空间上用的是邻接矩阵，结果超时。总结：这种方法适合于密集图。

后来考虑用dijsktra算法求单源最短路径，但是heap实现起来比较复杂，还得多个时间复杂度为O(logn)的decreasemin方法。网上看解题报告，看到了Shortest_Path_Faster_Algorithm。于是去wikipedia（http://en.wikipedia.org/wiki/Shortest_Path_Faster_Algorithm）上查看。思路很简单，据称最坏时间复杂 度非常坏，但在稀疏图中的时间复杂度好，为O(E)，并且可以应用在(无负环)的负数权值图中。于是照着伪代码实现了。效果非常好。之后照着wikipedia后面介绍的优化方法写了下，也很简单，也有时间上的提升。

TASK: butterLANG: C++Compiling...Compile: OKExecuting...   Test 1: TEST OK [0.000 secs, 5868 KB]   Test 2: TEST OK [0.000 secs, 5868 KB]   Test 3: TEST OK [0.000 secs, 5868 KB]   Test 4: TEST OK [0.000 secs, 5868 KB]   Test 5: TEST OK [0.000 secs, 5868 KB]   Test 6: TEST OK [0.000 secs, 5868 KB]   Test 7: TEST OK [0.011 secs, 5868 KB]   Test 8: TEST OK [0.022 secs, 5868 KB]   Test 9: TEST OK [0.043 secs, 5868 KB]   Test 10: TEST OK [0.043 secs, 5868 KB]All tests OK.

附注：一块草地上可以有不止一头牛。

代码如下：

/*ID: thestor1LANG: C++TASK: butter*/#include <iostream>#include <cmath>#include <cstdio>#include <cstring>#include <vector>#include <cassert>#include <string>#include <algorithm>#include <stack>#include <set>#include <queue>using namespace std;const int P = 800;const int infinity = 180000;int dis[P][P];struct Edge{int v;int w;Edge *next;Edge(int v, int w){this->v = v;this->w = w;this->next = NULL;}};struct Vertex{int u;Edge *adj;Vertex(int i){this->u = i;adj = NULL;}};void spfa(const int s, const int p, Vertex **vertexes){for(int i = 0; i < p; ++i){dis[s][i] = infinity;}dis[s][s] = 0;deque<int> que;bool onQueue[P];memset(onQueue, false, p * sizeof(bool));que.push_back(s);onQueue[s] = true;while(!que.empty()){int u = que.front();que.pop_front();onQueue[u] = false;for(Edge *e = vertexes[u]->adj; e != NULL; e = e->next){if(dis[s][u] + e->w < dis[s][e->v]){dis[s][e->v] = dis[s][u] + e->w;if(!onQueue[e->v]){if(que.empty() || dis[s][e->v] < dis[s][que.front()]){que.push_front(e->v);}else{que.push_back(e->v);}onQueue[e->v] = true;}}}}}int main(){FILE *fin  = fopen ("butter.in", "r");FILE *fout = fopen ("butter.out", "w");//freopen("log_1.txt", "w", stdout);int n, p, c;fscanf(fin, "%d%d%d", &n, &p, &c);Vertex **vertexes = new Vertex*[p];for(int i = 0; i < p; ++i){vertexes[i] = new Vertex(i);}vector<int> dests;for(int i = 0; i < n; ++i){int dest;fscanf(fin, "%d", &dest);dests.push_back(dest - 1);}for(int i = 0; i < c; ++i){int p1, p2, d;fscanf(fin, "%d%d%d", &p1, &p2, &d);Edge *euv = new Edge(p2 - 1, d);euv->next = vertexes[p1 - 1]->adj;vertexes[p1 - 1]->adj = euv;Edge *evu = new Edge(p1 - 1, d);evu->next = vertexes[p2 - 1]->adj;vertexes[p2 - 1]->adj = evu;}for(int i = 0; i < n; ++i){spfa(dests[i], p, vertexes);}int mindis = -1;for(int i = 0; i < p; ++i){int sum = 0;for(int j = 0; j < n; ++j){sum += dis[dests[j]][i];//fprintf(stdout, "%d->%d: %d, ", i, dests[j], dis[dests[j]][i]);}//fprintf(stdout, "\n%d: %d\n\n", i, sum);if(mindis < 0 || sum < mindis){mindis = sum;}}fprintf(fout, "%d\n", mindis);return 0;}