UVA 10246 - Asterix and Obelix（最短路）

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UVA 10246 - Asterix and Obelix

题目链接

题意：给定一个图，每个点有一个代价，边有一个代价，现在有q次询问，每次询问从u到v的最小花费，花费的计算方式为，路径代价加上路径上最大代价结点的代价

思路：枚举最大代价结点，然后做dijkstra，做的过程中忽略掉比枚举点更大代价的点，然后更新所有的答案，预处理完成后每次询问就可以在O（1）时间内完成了

代码：

#include <cstdio>#include <cstring>#include <vector>#include <queue>using namespace std;const int MAXNODE = 85;const int MAXEDGE = 10005;typedef int Type;const Type INF = 0x3f3f3f3f;struct Edge {int u, v;Type dist;Edge() {}Edge(int u, int v, Type dist) {this->u = u;this->v = v;this->dist = dist;}};struct HeapNode {Type d;int u;HeapNode() {}HeapNode(Type d, int u) {this->d = d;this->u = u;}bool operator < (const HeapNode& c) const {return d > c.d;}};int n, m, q, cost[MAXNODE], ans[MAXNODE][MAXNODE];struct Dijkstra {int n, m;Edge edges[MAXEDGE];int first[MAXNODE];int next[MAXEDGE];bool done[MAXNODE];Type d[MAXNODE];int p[MAXNODE];void init(int n) {this->n = n;memset(first, -1, sizeof(first));m = 0;}void add_Edge(int u, int v, Type dist) {edges[m] = Edge(u, v, dist);next[m] = first[u];first[u] = m++;}void dijkstra(int s) {priority_queue<HeapNode> Q;for (int i = 0; i < n; i++) d[i] = INF;d[s] = 0;p[s] = -1;memset(done, false, sizeof(done));Q.push(HeapNode(0, s));while (!Q.empty()) {HeapNode x = Q.top(); Q.pop();int u = x.u;if (done[u]) continue;done[u] = true;for (int i = first[u]; i != -1; i = next[i]) {Edge& e = edges[i];if (cost[e.v] > cost[s]) continue;if (d[e.v] > d[u] + e.dist) {d[e.v] = d[u] + e.dist;p[e.v] = i;Q.push(HeapNode(d[e.v], e.v));}}}for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {ans[i][j] = min(ans[i][j], d[i] + d[j] + cost[s]);}}}} gao;int main() {int cas = 0;int bo = 0;while (~scanf("%d%d%d", &n, &m, &q) && n) {if (bo) printf("\n");else bo = 1;gao.init(n);for (int i = 0; i < n; i++) scanf("%d", &cost[i]);int u, v, d;while (m--) {scanf("%d%d%d", &u, &v, &d);u--; v--;gao.add_Edge(u, v, d);gao.add_Edge(v, u, d);}memset(ans, INF, sizeof(ans));for (int i = 0; i < n; i++) gao.dijkstra(i);for (int i = 0; i < n; i++)for (int j = 0; j < n; j++)if (ans[i][j] == INF) ans[i][j] = -1;printf("Case #%d\n", ++cas);while (q--) {scanf("%d%d", &u, &v);u--; v--;printf("%d\n", ans[u][v]);}}return 0;}

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