Heavy Transportation POJ - 1797
2017-10-09 15:58
405 查看
Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has build his giant steel crane to the place where it is needed on which all streets can carry the weight. Fortunately he already has a plan of the city with all streets and bridges and all the allowed weights.Unfortunately he has no idea how to find the the maximum weight capacity in order to tell his customer how heavy the crane may become. But you surely know. Problem You are given the plan of the city, described by the streets (with weight limits) between the crossings, which are numbered from 1 to n. Your task is to find the maximum weight that can be transported from crossing 1 (Hugo's place) to crossing n (the customer's place). You may assume that there is at least one path. All streets can be travelled in both directions.
Input
The first line contains the number of scenarios (city plans). For each city the number n of street crossings (1 <= n <= 1000) and number m of streets are given on the first line. The following m lines contain triples of integers specifying start and end crossing of the street and the maximum allowed weight, which is positive and not larger than 1000000. There will be at most one street between each pair of crossings.
Output
The output for every scenario begins with a line containing “Scenario #i:”, where i is the number of the scenario starting at 1. Then print a single line containing the maximum allowed weight that Hugo can transport to the customer. Terminate the output for the scenario with a blank line.
Sample Input
1 3 3 1 2 3 1 3 4 2 3 5
Sample Output
Scenario #1: 4
题意:求从1到n的最短路的最大权(优先最短路,然后取权值最大的)
/*找到一条从1到n的路径,使得路上的最小边权最大*/ #include<stdio.h> #include<math.h> #include<string.h> #define N 1100 #define Min(a,b) a<b?a:b #define Max(a,b) a>b?a:b int map ; int vis ,dis ; int n,m; int dijkstra() { int i,j,pos,max; memset(vis,0,sizeof(vis)); for(i=1;i<=n;i++) { dis[i] = map[1][i]; } vis[1] =1; for(i=2;i<=n;i++) { max = -1; for(j=1;j<=n;j++) { if(!vis[j]&&dis[j]>max) { max = dis[j]; pos = j; } } if(max == -1) break; vis[pos] = 1; for(j=1;j<=n;j++) { dis[j] = Max(Min(dis[pos],map[pos][j]),dis[j]); } } } int main() { int i,k,t; int a,b,c; scanf("%d",&t); for(k = 1;k<=t;k++) { scanf("%d %d",&n,&m); memset(map,0,sizeof(map)); for(i=1;i<=m;i++) { scanf("%d %d %d",&a,&b,&c); map[a][b] = map[b][a] = c; } printf("Scenario #%d:\n",k); dijkstra(); printf("%d\n\n",dis ); } return 0; }
相关文章推荐
- POJ 1797 ——Heavy Transportation——————【最短路、Dijkstra、最短边最大化】
- 最短路 - C - Heavy Transportation POJ - 1797
- Heavy Transportation POJ - 1797
- Heavy Transportation POJ - 1797
- Heavy Transportation POJ - 1797
- 【最短路入门专题1】E - Heavy Transportation Poj 1797【Dijkstra变形题】
- POJ 1797 Heavy Transportation / SCU 1819 Heavy Transportation (图论,最短路径)
- Heavy Transportation POJ - 1797
- Heavy Transportation POJ - 1797 --- 最短路思维的最大承载量(最大生成树)
- POJ 1797 Heavy Transportation Dijstr最短路变形
- POJ 1797 Heavy Transportation and POJ - 2253 Frogger【边权最大问题】
- POJ 1797 Heavy Transportation&&POJ 2253 Frogger 最短路 dijkstra变形
- POJ 1797 Heavy Transportation&&POJ 2253 Frogger 最短路 dijkstra变形
- poj Heavy Transportation
- poj 1797 Heavy Transportation
- [POJ 1797]Heavy Transportation[SPFA]
- poj 1797 Heavy Transportation
- POJ1797 Heavy Transportation(SPFA)
- POJ1797 Heavy Transportation
- POJ 1797 图论 Dijkstra