POJ3268: Silver Cow Party

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本文介绍了一个关于寻找从多个起点到指定终点再返回起点的最短路径问题，并提供了一个具体的实现案例，通过Dijkstra算法计算出所有牛从各自农场到派对地点再返回所需的最大时间。

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Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5738		Accepted: 2508

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 T_i (1 ≤ T_i ≤ 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: A_i, B_i, and T_i. The described road runs from farm A_i to farm B_i, requiring T_i time units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

Sample Input

Sample Output

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

USACO 2007 February Silver

解题思路：典型的最短路，我写过dijkstra的模板，所以只是复制过来改改代码就可以了

PS：太TM贱了，经过测试，T的最大值绝对超过1000，而非题目中说的1≤T≤100，我一开始把infi设置为120，W.A.了。

//#include <windows.h>
//#include <iostream>
//#include <fstream>
#include <cstdio>
//#include <cmath>
//#include <cstdlib>
//#include <cstring>
//#include <algorithm>
//using namespace std;

//LARGE_INTEGER _lTemp;
//LONGLONG _dFreq, _bTime, _eTime;
//double _uTime;

const int infi = 10000000;
int size; 	//顶点数
int edge[1020][1020];	//edge[i][j]保存Vi→Vj权值
int distIn[1020], distOt[1020];
bool setIn[1020], setOt[1020];
int main(){
//	freopen("in.txt", "r", stdin);	//重定向输入
//	freopen("out.txt", "w", stdout);//重定向输出
//	QueryPerformanceFrequency(&_lTemp);							//获取运算频率
//	_dFreq = _lTemp.QuadPart;									//获取运算频率大整数
//	QueryPerformanceCounter(&_lTemp);							//获取开始时间
//	_bTime = _lTemp.QuadPart;									//获取开始时间大整数

int m, tar, x, y, t, max = 0;
	scanf("%d%d%d", &size, &m, &tar);
	for (int i = 1; i <= size; i ++)
		for (int j = 1; j <= size; j ++)
			edge[i][j] = (i == j ? 0 : infi);
	while (m --){
		scanf("%d%d%d", &x, &y, &t);
		if (edge[x][y] > t)
			edge[x][y] = t;
	}
	for (int i = 1; i <= size; i++){
		distIn[i] = edge[i][tar];
		distOt[i] = edge[tar][i];
		setIn[i] = setOt[i] = false;
	}
	setIn[tar] = setOt[tar] = true;
	for (int i = 0; i < size - 1; i ++){
		int min = infi, u = tar;
		for (int j = 1; j <= size; j ++)
			if (!setIn[j] && distIn[j] < min)
			{ u = j; min = distIn[j]; }
		setIn[u] = true;
		for (int j = 1; j <= size; j ++)
			if (!setIn[j] && edge[j][u] < infi && distIn[u] + edge[j][u] < distIn[j])
				distIn[j] = distIn[u] + edge[j][u];
	}
	for (int i = 0; i < size - 1; i ++){
		int min = infi, u = tar;
		for (int j = 1; j <= size; j ++)
			if (!setOt[j] && distOt[j] < min)
			{ u = j; min = distOt[j]; }
		setOt[u] = true;
		for (int j = 1; j <= size; j ++)
			if (!setOt[j] && edge[u][j] < infi && distOt[u] + edge[u][j] < distOt[j])
				distOt[j] = distOt[u] + edge[u][j];
	}
	for (int i = 1; i <= size; i ++)
		if (distIn[i] + distOt[i] > max)
			max = distIn[i] + distOt[i];
	printf("%d/n", max);

//	QueryPerformanceCounter(&_lTemp);							//获取终止时间
//	_eTime = _lTemp.QuadPart;									//获取终止时间大整数
//	_uTime = 1000 * double(_eTime - _bTime) / double(_dFreq);	//计算总耗时，转化成ms
//	printf("Time used: %012.3f ms/n", _uTime);					//输出总耗时，单位是ms
//	fclose(stdin);	//关闭文件
//	fclose(stdout);	//关闭文件

return 0;
}