HDU 1598 find the most comfortable road
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find the most comfortable road
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8072 Accepted Submission(s): 3394
Problem Description
XX星有许多城市,城市之间通过一种奇怪的高速公路SARS(Super Air Roam Structure---超级空中漫游结构)进行交流,每条SARS都对行驶在上面的Flycar限制了固定的Speed,同时XX星人对 Flycar的“舒适度”有特殊要求,即乘坐过程中最高速度与最低速度的差越小乘坐越舒服 ,(理解为SARS的限速要求,flycar必须瞬间提速/降速,痛苦呀 ),
但XX星人对时间却没那么多要求。要你找出一条城市间的最舒适的路径。(SARS是双向的)。
但XX星人对时间却没那么多要求。要你找出一条城市间的最舒适的路径。(SARS是双向的)。
Input
输入包括多个测试实例,每个实例包括:
第一行有2个正整数n (1<n<=200)和m (m<=1000),表示有N个城市和M条SARS。
接下来的行是三个正整数StartCity,EndCity,speed,表示从表面上看StartCity到EndCity,限速为speedSARS。speed<=1000000
然后是一个正整数Q(Q<11),表示寻路的个数。
接下来Q行每行有2个正整数Start,End, 表示寻路的起终点。
第一行有2个正整数n (1<n<=200)和m (m<=1000),表示有N个城市和M条SARS。
接下来的行是三个正整数StartCity,EndCity,speed,表示从表面上看StartCity到EndCity,限速为speedSARS。speed<=1000000
然后是一个正整数Q(Q<11),表示寻路的个数。
接下来Q行每行有2个正整数Start,End, 表示寻路的起终点。
Output
每个寻路要求打印一行,仅输出一个非负整数表示最佳路线的舒适度最高速与最低速的差。如果起点和终点不能到达,那么输出-1。
Sample Input
4 41 2 22 3 41 4 13 4 221 31 2
Sample Output
10
Author
ailyanlu
Source
HDU 2007-Spring Programming Contest - Warm Up (1)
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按边长从小到大排序,然后从小到枚举边长,以该边作为最小值,再将比它大的边逐渐合并如果start与end的根节点相同说明联通,该边产生的最小差就是lenj - leni;
#include <algorithm>#include <iostream>#include <cstring>#include <cstdio>#include <queue>using namespace std;#define INF 0x3f3f3f3f#define N 1100struct node { int u,v; int w;}edge[N*5];int father[N];int n, m;bool cmp1(node a, node b){ return a.w < b.w;}int findfather(int x){ int a = x; while(x != father[x]) x = father[x]; while(a != father[a]) { int t = a; a = father[a]; father[t] = x; } return x;}void init(){ for(int i = 0; i <= n; i++) father[i] = i; return ;}int main(){ while(scanf("%d%d", &n, &m) != EOF) { for(int i = 0; i < m; i++) scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].w); sort(edge, edge + m, cmp1); //边长从小到大排序 int Start, End ,q; int ans = INF; scanf("%d", &q); for(int k = 0; k < q; k++) { scanf("%d%d", &Start, &End); ans = INF; for(int i = 0; i < m; i++) { init(); for(int j = i; j < m; j++) { int faA = findfather(edge[j].u); int faB = findfather(edge[j].v); father[faB] = faA; if(findfather(Start) == findfather(End)) { //说明已经联通 ans = min(ans, edge[j].w - edge[i].w); //因为边长从小到大排序 break; } } } if(ans == INF) printf("-1\n"); else printf("%d\n", ans); } } return 0;}
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