POJ 3268 Silver Cow Party 单向最短路
2024-08-31 10:42:57
Silver Cow Party
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 22864 | Accepted: 10449 |
Description
One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires Ti (1 ≤ Ti ≤ 100) units of time to traverse.
Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.
Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?
Input
Line 1: Three space-separated integers, respectively: N, M, and X
Lines 2..M+1: Line i+1 describes road i with three space-separated integers: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.
Lines 2..M+1: Line i+1 describes road i with three space-separated integers: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.
Output
Line 1: One integer: the maximum of time any one cow must walk.
Sample Input
4 8 2
1 2 4
1 3 2
1 4 7
2 1 1
2 3 5
3 1 2
3 4 4
4 2 3
Sample Output
10
Hint
Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.
Source
求两次最短路,第一次求t到其余各点的最短路,第二次求各点到t的最短路。
最后求的是所有点到t的距离的最大值
#include <iostream>
#include <deque>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
#define maxn 100010
using namespace std;
vector<pair<int,int> > E[maxn];
int d1[maxn],d2[maxn];
int n,m,t,x[maxn],y[maxn],z[maxn];
int max(int a,int b){
if(a>=b){
return a;
}
return b;
}
void init1(){
for(int i=;i<maxn;i++){
E[i].clear();
d1[i] = 1e9;
}
}
void init2(){
for(int i=;i<maxn;i++){
E[i].clear();
d2[i] = 1e9;
}
}
void dijkstra(int t,int d[]){
d[t] = ;
priority_queue<pair<int,int> > q;
q.push(make_pair(-d[t],t));
while(!q.empty()){
int now = q.top().second;
q.pop();
for(int i=;i<E[now].size();i++){
int v = E[now][i].first;
if(d[v] > d[now] + E[now][i].second){
d[v] = d[now] + E[now][i].second;
q.push(make_pair(-d[v],v));
}
}
}
}
int main()
{
while(cin >> n >> m >> t){
init1();
for(int i=;i<m;i++){
cin >> x[i] >> y[i] >> z[i];
E[x[i]].push_back(make_pair(y[i],z[i]));
}
dijkstra(t,d1);//正求一次
init2();
for(int i=;i<m;i++){
E[y[i]].push_back(make_pair(x[i],z[i]));
} //记得在这里要把所有的路反过来
dijkstra(t,d2);//反求一次
int num = -;
for(int i=;i<=n;i++){
num = max(num,d1[i]+d2[i]);
}
cout << num << endl;
}
return ;
}
最新文章
- LPC1769 CAN的自测试模式
- android中 onResume()方法什么时候执行 ??(转)
- Android--RecyclerView的封装使用
- 输入输出流(IO)
- C++11的default和delete关键字
- dSYM atos crash log 定位到代码行的方法(转)
- 如何在nopcommerce3.3注册页面添加密码强度检查仪?
- Yii2 如何使用事件
- 【spring】non-compatible bean definition of same name and class
- web服务器分析与设计(四)
- 深入分析 Java I/O 的工作机制--转载
- Delphi中文本文件的操作
- 1.4 random模块
- python一直放弃到双数的day10
- C/C++的20个位运算技巧
- 简单Hash函数LongHash
- 本地项目提交到github和提交更新(转)
- C++对windows控制面板的操作
- [LeetCode] 236. Lowest Common Ancestor of a Binary Tree_ Medium tag: DFS, Divide and conquer
- 《剑指offer》总结二 之二叉树