poj 1797
2017-03-08 21:37
225 查看
Heavy Transportation
Description
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
Sample Output
题解:求从原点到终点的最大承重量,单源路径问题,可以根据迪杰斯特拉算法,这时首先要找dis数组中最大的,以它为起点,更新
dis数组,要比较max和map[flag][j]最小的,然后再取和dis[j]中最大的,即在同一条路中选择最小的,不同路中选择最大的。
#include<stdio.h>
#define INF 1<<30
int dis[1001], weight[1001][1001], visited[1001] = {0};
void get_weight(int weight[][1001], int n) {
int i, j, max, flag;
for(i = 1; i <= n; i++)
dis[i] = weight[1][i];
visited[1] = 1;
for(i = 1; i < n; i++) {
max = -INF;
for(j = 1; j <= n; j++)
if(dis[j] > max && !visited[j]) {
max = dis[j];
flag = j;
}
visited[flag] = 1;
for(j = 1; j <= n; j++) {
if(dis[j] < (max > weight[flag][j] ? weight[flag][j] : max) && !visited[j])
dis[j] = max > weight[flag][j] ? weight[flag][j] : max;
}
}
}
int main() {
int t, n, m,count = 1;
scanf("%d", &t);
while(t--) {
int i, j, street1, street2, wei;
for(i = 0; i < 1001; i++) {
visited[i] = 0;
for(j = 0; j < 1001; j++)
weight[i][j] = -1;
}
scanf("%d%d", &n, &m);
for(i = 1; i <= m; i++) {
scanf("%d%d%d", &street1, &street2, &wei);
if(wei > weight[street1][street2]) {
weight[street1][street2] = wei;
weight[street2][street1] = wei;
}
}
get_weight(weight, n);
printf("Scenario #%d:\n", count);
count++;
printf("%d\n", dis
);
printf("\n");
}
}
Time Limit: 3000MS | Memory Limit: 30000K | |
Total Submissions: 33180 | Accepted: 8803 |
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
题解:求从原点到终点的最大承重量,单源路径问题,可以根据迪杰斯特拉算法,这时首先要找dis数组中最大的,以它为起点,更新
dis数组,要比较max和map[flag][j]最小的,然后再取和dis[j]中最大的,即在同一条路中选择最小的,不同路中选择最大的。
#include<stdio.h>
#define INF 1<<30
int dis[1001], weight[1001][1001], visited[1001] = {0};
void get_weight(int weight[][1001], int n) {
int i, j, max, flag;
for(i = 1; i <= n; i++)
dis[i] = weight[1][i];
visited[1] = 1;
for(i = 1; i < n; i++) {
max = -INF;
for(j = 1; j <= n; j++)
if(dis[j] > max && !visited[j]) {
max = dis[j];
flag = j;
}
visited[flag] = 1;
for(j = 1; j <= n; j++) {
if(dis[j] < (max > weight[flag][j] ? weight[flag][j] : max) && !visited[j])
dis[j] = max > weight[flag][j] ? weight[flag][j] : max;
}
}
}
int main() {
int t, n, m,count = 1;
scanf("%d", &t);
while(t--) {
int i, j, street1, street2, wei;
for(i = 0; i < 1001; i++) {
visited[i] = 0;
for(j = 0; j < 1001; j++)
weight[i][j] = -1;
}
scanf("%d%d", &n, &m);
for(i = 1; i <= m; i++) {
scanf("%d%d%d", &street1, &street2, &wei);
if(wei > weight[street1][street2]) {
weight[street1][street2] = wei;
weight[street2][street1] = wei;
}
}
get_weight(weight, n);
printf("Scenario #%d:\n", count);
count++;
printf("%d\n", dis
);
printf("\n");
}
}
相关文章推荐
- POJ - 1797----Heavy Transportation(Dijkstra)
- POJ - 1797 最短路思想加以变化解此题..
- POJ 1797 Heavy Transportation
- POJ1797---Heavy Transportation(最小生成树变形求最长路)
- POJ - 1797 Heavy Transportation(dijkstra最短路求最大边)
- poj 1797(最短路变形)
- POJ-1797-Heavy Transportation(草稿)
- POJ 1797 Heavy Transportation 笔记
- POJ - 1797 Heavy Transportation
- POJ 1797 Heavy Transportation Kruskal并查集求解
- [kuangbin带你飞]专题四 最短路练习 C POJ 1797
- poj 1797
- poj1797 Heavy Transportation
- poj 1797 Heavy transportation 最短路
- poj1797 - Heavy Transportation
- poj1797
- Dijkstra||Prim-POJ-1797-Heavy Transportation
- POJ 1797 Heavy Transportation
- POJ 1797 Heavy Transportion Dijkstra变形
- poj1797 Heavy Transportation(最短路变形)