BZOJ 1631==USACO 2007== POJ 3268 Cow Party奶牛派对

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 19226		Accepted: 8775

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.

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int map[1001][1001],w[1001][1001];
int dis[1001],Team[5000];
bool exist[1001];
int n,m,t;        // exist 为 true 则不可进队
void SPFA(int s)
{
int head=0,tail=1,k=0;
Team[head]=s,exist[s]=true;dis[s]=0;
while(head<tail)
{
k=Team[head];
exist[k]=false;
for(int i=1;i<=n;i++)
{
if((map[k][i]>0)&&(dis[i]>dis[k]+map[k][i]))
{
dis[i]=map[k][i]+dis[k];
if(exist[i]==false)
{
Team[tail++]=i;
exist[i]=true;
}
}
}
head++;
}
}
int main()
{
cin>>n>>m>>t;
memset(map,0x3f,sizeof map );
memset(w,0x3f,sizeof w );
memset(dis,0x3f,sizeof dis );
for(int i=1;i<=m;i++)
{
int x,y,z;
cin>>x>>y>>z;
map[x][y]=z;
w[x][y]=z;
}
SPFA(t);// 从起点开始到其他每个点找一遍最短路
for(int k=1;k<=n;k++)
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(w[i][j]>w[i][k]+w[k][j])
w[i][j]=w[i][k]+w[k][j];
}
}
}
int max;
for(int i=1;i<=n;i++)
{
if((i==1)||(dis[i]+w[i][t]>max))
max=dis[i]+w[i][t];
}
printf("%d",max);
return 0;
}

思路：一遍SPFA求出起点到其他所有点的最短路，一遍Floyed求出所有点到起点的最短路，两者相加，和最大者即为anwser...

BZOJ 1631==USACO 2007== POJ 3268 Cow Party奶牛派对相关推荐

USACO 2.3.2 Cow Pedigrees 奶牛家谱
题目描述: 农民约翰准备购买一群新奶牛. 在这个新的奶牛群中, 每一个母亲奶牛都生两个小奶牛.这些奶牛间的关系可以用二叉树来表示.这些二叉树总共有N个节点(3 <= N < 200).这些 ...
[Usaco2007 Feb]Cow Party 奶牛派对
好了,好久没发题解,这次来一发SPFA,看题: Description 农场有N(1≤N≤1000)个牛棚,每个牛棚都有1只奶牛要参加在X牛棚举行的奶牛派对．共有M(1≤M≤100000)条单向路连 ...
poj 3268 bzoj 1631: [Usaco2007 Feb]Cow Party（最短路）
1631: [Usaco2007 Feb]Cow Party Time Limit: 5 Sec Memory Limit: 64 MB Submit: 855 Solved: 613 [Subm ...
D - Silver Cow Party POJ - 3268
D - Silver Cow Party POJ - 3268 dijkstra 是 O(n2),堆优化一下, O(nlogn) 对每个点跑一次 dj, 取 max(dis(x->i)+dis( ...
POJ 3268 Silver Cow Party （最短路径）
POJ 3268 Silver Cow Party (最短路径) Description One cow from each of N farms (1 ≤ N ≤ 1000) convenientl ...
Silver Cow Party （POJ - 3268 ）
Silver Cow Party (POJ - 3268 ) 这道题是我做的最短路专题里的一道题,但我还没做这个,结果比赛就出了,真是.......... 题目: One cow from each ...
POJ 3263-Tallest Cow
Description FJ's N (1 ≤ N ≤ 10,000) cows conveniently indexed 1..N are standing in a line. Each cow ...
POJ 3268 Bookshelf 2 动态规划法题解
Description Farmer John recently bought another bookshelf for the cow library, but the shelf is gett ...
[USACO 2007 Jan S]Protecting the Flowers
题目: [USACO 2007 Jan S]Protecting the Flowers ,哈哈,我们今天来看一道简单的贪心算法题嘛,这是选自USACO上的一道题,好了,我们一起来看看题意吧: 题目描 ...

BZOJ 1631==USACO 2007== POJ 3268 Cow Party奶牛派对

BZOJ 1631==USACO 2007== POJ 3268 Cow Party奶牛派对相关推荐

最新文章

热门文章