Language:
Out of Hay
Time Limit:1000MSMemory Limit:65536K
Total Submissions:11842Accepted:4639
Description
The cows have run out of hay, a horrible event that must be remedied immediately. Bessie intends to visit the other farms to survey their hay situation. There are N (2 <= N <= 2,000) farms (numbered 1..N); Bessie starts at Farm 1. She’ll traverse some or all of the M (1 <= M <= 10,000) two-way roads whose length does not exceed 1,000,000,000 that connect the farms. Some farms may be multiply connected with different length roads. All farms are connected one way or another to Farm 1.Bessie is trying to decide how large a waterskin she will need. She knows that she needs one ounce of water for each unit of length of a road. Since she can get more water at each farm, she’s only concerned about the length of the longest road. Of course, she plans her route between farms such that she minimizes the amount of water she must carry.Help Bessie know the largest amount of water she will ever have to carry: what is the length of longest road she’ll have to travel between any two farms, presuming she chooses routes that minimize that number? This means, of course, that she might backtrack over a road in order to minimize the length of the longest road she’ll have to traverse.
Input
* Line 1: Two space-separated integers, N and M.* Lines 2..1+M: Line i+1 contains three space-separated integers, A_i, B_i, and L_i, describing a road from A_i to B_i of length L_i.
Output
* Line 1: A single integer that is the length of the longest road required to be traversed.
Sample Input
3 31 2 232 3 10001 3 43
Sample Output
43
Hint
OUTPUT DETAILS:In order to reach farm 2, Bessie travels along a road of length 23. To reach farm 3, Bessie travels along a road of length 43. With capacity 43, she can travel along these roads provided that she refills her tank to maximum capacity before she starts down a road.
Source
题意：求最小生成树的最大边。
复习了prim算法和kruskal算法。
代码：
//Kruskal#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 10105#define MAXN 2005#define mod 1000000009#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,r#define FRE(i,a,b) for(i = a; i <= b; i++)#define FREE(i,a,b) for(i = a; i >= b; i–)#define FRL(i,a,b) for(i = a; i < b; i++)#define FRLL(i,a,b) for(i = a; i > b; i–)#define mem(t, v) memset ((t) , v, sizeof(t))#define sf(n)scanf("%d", &n)#define sff(a,b) scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pfprintf#define DBGpf("Hi\n")typedef long long ll;using namespace std;struct Edge{int u,v,w;}edge[maxn];int father[2222];int n,m;int cmp(Edge a,Edge b){return a.w<b.w;}int find_father(int x){if (x!=father[x])father[x]=find_father(father[x]);return father[x];}void Kruskal(){int i;FRL(i,0,n+10)father[i]=i;int cnt=0;int ans=-1;FRL(i,0,m){int fa=find_father(edge[i].u);int fb=find_father(edge[i].v);if (fa!=fb){father[fa]=fb;ans=max(ans,edge[i].w);cnt++;}if (cnt==n-1) break;}pf("%d\n",ans);}int main(){int i,j;while (~sff(n,m)){FRL(i,0,m)sfff(edge[i].u,edge[i].v,edge[i].w);sort(edge,edge+m,cmp);Kruskal();}return 0;}//prim#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 2105#define MAXN 2005#define mod 1000000009#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,r#define FRE(i,a,b) for(i = a; i <= b; i++)#define FREE(i,a,b) for(i = a; i >= b; i–)#define FRL(i,a,b) for(i = a; i < b; i++)#define FRLL(i,a,b) for(i = a; i > b; i–)#define mem(t, v) memset ((t) , v, sizeof(t))#define sf(n)scanf("%d", &n)#define sff(a,b) scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pfprintf#define DBGpf("Hi\n")typedef long long ll;using namespace std;int mp[maxn][maxn];int n,m;int dist[maxn];int vis[maxn];void prim(){int i,j,maxx=-1,now;dist[1]=0;FRL(i,0,n){now=-1;FRE(j,1,n){if ( !vis[j] && ( now==-1 || dist[j]<dist[now] ))now=j;}if (now==-1) return ;maxx=max(maxx,dist[now]);vis[now]=1;FRE(j,1,n)if (!vis[j]&&dist[j]>mp[now][j])dist[j]=mp[now][j];}pf("%d\n",maxx);return ;}int main(){int i,j;while (~sff(n,m)){int u,v,w;FRL(i,0,n+10){FRL(j,0,n+10){if (i==j)mp[i][j]=0;elsemp[i][j]=INF;}vis[i]=0;dist[i]=INF;}FRL(i,0,m){sfff(u,v,w);if (mp[u][v]>w)mp[u][v]=mp[v][u]=w;}prim();}return 0;}
，当你能飞的时候就不要放弃飞
