HDU 1690 Bus System

最新推荐文章于 2020-01-10 19:05:31 发布

原创最新推荐文章于 2020-01-10 19:05:31 发布 · 975 阅读

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HDU-2010 summer（1）最短路专题专栏收录该内容

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本文介绍了一种使用SPFA算法解决图中单源最短路径问题的方法，并通过具体实例展示了如何根据节点间的距离设置不同的权重成本，进而求解任意两点间的最小花费。特别注意到了在特定情况下，路径长度可能超过整型变量的表示范围，因此采用了合适的数据类型来避免溢出问题。

题目比较简单，根据图给的方式建边（len->cost）。

唯一需要注意的地方：each number is non-negative and not larger than 1,000,000,000.但是最多是100个点，

路径最长的时候10^9 * (100 - 1)已经超出int范围了。

00446728

2010-07-04 19:27:18

Accepted

1013

171 MS

304 KB

Visual C++

Slave_wc

#include <iostream> #include <cstdio> #include <queue> #include <cmath> #include <cstring> #define MAXN 120 #define INF 0x7f7f7f7f7f7f7f7fLL using namespace std; __int64 map[MAXN][MAXN]; __int64 dis[MAXN]; bool hash[MAXN]; struct Dij { int id; __int64 dis; bool friend operator < (const Dij a, const Dij b) { return a.dis > b.dis; } }; Dij now, next; __int64 SPFA(int st, int ed, int n) { memset(hash, false, sizeof(hash)); memset(dis, 0x7f, sizeof(dis)); priority_queue<Dij> Q; now.id = st; now.dis = 0; dis[st] = 0; Q.push(now); while (!Q.empty()) { now = Q.top(); Q.pop(); if (hash[now.id]) { continue; } if (now.id == ed) { return now.dis; } hash[now.id] = true; for (int j = 1; j <= n; j++) { if (!hash[j] && dis[j] > dis[now.id] + map[now.id][j]) { dis[j] = dis[now.id] + map[now.id][j]; next.id = j; next.dis = dis[j]; Q.push(next); } } } return -1; } int main() { int test = 1; int t, i, j, n, m, cost[5], len[5], x[MAXN]; scanf("%d", &t); while (t--) { for (i = 1; i <= 4; i++) { scanf("%d", &len[i]); } for (i = 1; i <= 4; i++) { scanf("%d", &cost[i]); } scanf("%d %d", &n, &m); for (i = 1; i <= n; i++) { scanf("%d", &x[i]); } for (i = 1; i <= n; i++) { for (j = i + 1; j <= n; j++) { int dis = (int)abs(x[i] - x[j]); if (dis > len[4]) { map[i][j] = map[j][i] = INF; } else if (dis > len[3]) { map[i][j] = map[j][i] = cost[4]; } else if (dis > len[2]) { map[i][j] = map[j][i] = cost[3]; } else if (dis > len[1]) { map[i][j] = map[j][i] = cost[2]; } else { map[i][j] = map[j][i] = cost[1]; } } } printf("Case %d:/n", test++); for (i = 0; i < m; i++) { int st, ed; __int64 ans; scanf("%d %d", &st, &ed); ans = SPFA(st, ed, n); if (ans == -1) { printf("Station %d and station %d are not attainable./n", st, ed); } else { printf("The minimum cost between station %d and station %d is %I64d./n", st, ed, ans); } } } return 0; }