最短路径变形（最大值的最小值）

http://poj.org/problem?id=1797

题意：n个城市，m条双向边每条边有个承受重量权值，问1到n最大承重量。

解法：初始化为0，选权值大的边，更新：如果dis[j] < min(dis[pos] , ma[pos][j]) 这样更新使得到达j的承重量尽可能的大。

//#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
#include <stdlib.h>
using namespace std;
typedef long long ll ;
#define int ll
#define mod 100
#define gcd(m,n) __gcd(m, n)
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j)  for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//int lcm(int a , int b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define INF  0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define pb push_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define size(v) (int)(v.size())
#define cin(x) scanf("%lld" , &x);
const int N = 1e5+9;
const int maxn = 1e3+9;
const double esp = 1e-6;
int ma[maxn][maxn] , vis[maxn] , dis[maxn];
int n , m , cnt;

void dijkstra(int u){
    rep(i , 1 , n){
        dis[i] = ma[u][i];
    }
    vis[u] = 1 ;
    rep(i , 1 , n-1){
        int pos ;
        int maxx = -INF;
        rep(j , 1 , n){
            if(!vis[j] && maxx < dis[j]){
                pos = j ;
                maxx = dis[j];
            }
        }
        vis[pos] = 1 ;
        rep(j , 1 , n){
            if(!vis[j] && min(dis[pos] , ma[pos][j]) > dis[j]){
                dis[j] = min(dis[pos] , ma[pos][j]);
            }
        }
    }
}

void init(){
    ME(ma , 0);
    ME(vis , 0);
}
void solve(){
    init();
    scanf("%lld%lld" , &n , &m);
    rep(i , 1 , m){
        int u , v , w ;
        scanf("%lld%lld%lld" , &u , &v , &w);
        ma[u][v] = ma[v][u] = max(ma[v][u] , w);
    }
    dijkstra(1);
    cout << "Scenario #" << ++cnt << ":" << endl << dis[n] << endl << endl;
}

signed main()
{
    init();
    int t ;
    cin >> t ;
    while(t--){
        solve();
    }
}

 http://poj.org/problem?id=2253

题意：一只青蛙要跳到另一只青蛙的所在石头，给出n个石头，从1到n，使跳跃的所需最小距离范围最小为多少？

解法：更新：如果dis[j] > max(dis[pos] , pos[j]) , 使得到达dis【j】的最长距离最小

//#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
#include <stdlib.h>
using namespace std;
typedef long long ll ;
#define int ll
#define mod 1000000007
#define gcd(m,n) __gcd(m, n)
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j)  for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
int lcm(int a , int b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define INF  0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define pb push_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define size(v) (int)(v.size())
#define cin(x) scanf("%lld" , &x);
const int N = 1e7+9;
const int maxn = 2e2+9;
const double esp = 1e-6;
double ma[maxn][maxn];
double dis[maxn];
int vis[maxn], n , m, cnt;
pii a[maxn];

void dijia(int r){
    rep(i , 1 , n){
        dis[i] = ma[r][i];
    }
    vis[r] = 1 ;
    rep(i , 1 , n-1){
        int pos ;
        double mi = INF;
        rep(j , 1 , n){
            if(!vis[j] && mi > dis[j]){
                pos = j ;
                mi = dis[j];
            }
        }
        vis[pos] = 1 ;
        rep(j , 1 , n){
            double x = max(dis[pos] , ma[pos][j]);
            if(!vis[j] && x < dis[j]){
                dis[j] = x ;
            }
        }
    }
}
void init(){
    fill(ma[0] , ma[0]+maxn*maxn , INF);
    ME(vis,0);
}
void solve(){
    init();
    int bx , by , ex , ey;
    scanf("%lld%lld%lld%lld" , &bx , &by , &ex , &ey);
    a[1].fi = bx , a[1].se = by , a[n].fi = ex , a[n].se = ey;
    rep(i , 2 , n-1){
        int u , v ;
        scanf("%lld%lld" , &a[i].fi , &a[i].se);
    }
    rep(i , 1 , n){
        rep(j , 1 , i-1){
            ma[i][j] = ma[j][i] = min(ma[i][j] , sqrt((a[i].fi - a[j].fi)*(a[i].fi - a[j].fi) + (a[i].se - a[j].se) * (a[i].se - a[j].se)));
        }
    }
    dijia(1);
    cout << "Scenario #" << ++cnt << endl;
    printf("Frog Distance = %.3f

" , dis[n]);
}

signed main()
{
    //int t ;
    //cin >> t ;
    //while(t--)
    while(~scanf("%lld" , &n) && n)
        solve();
}