Poj3662 Usaco2008JanSilver Telephone Lines

题目大意
给定N个点P条边的无向图，官方提供k条免费边。求如何去掉免费边，使剩下的边之中最大的边权最小，输出最大的边权

这题…最大的最小，很容易想到二分…而且这些边一定构成了最短路。
啊啊啊啊啊，我的代码在洛谷上过了，但是在poj上死活不过=。=

#include <cstdio>#include <queue>#include <iostream>#include <algorithm>#include <cstring>const int maxn = 100100;int n,m,begin,end,k,au[maxn],av[maxn],aw[maxn],diss[maxn],ans,r;struct edge{    int u,v,w,next,ww;} e[maxn];int a[maxn],num;inline void addedge(int u,int v,int w){    num++;    e[num].u = u;    e[num].v = v;    e[num].w = w;    e[num].next = a[u];    a[u] = num;}int dis[maxn],vis[maxn];struct st{    int n,w;    bool operator <(const st &a) const{        return a.w < w;    }};std::priority_queue<st> q;void dijkstra(int l){    memset(vis,0,sizeof(vis));    for(int i = 1;i <= n;i++)        dis[i] = 0x7fffffff;    dis[begin] = 0;    q.push((st){begin,0});    while(!q.empty()){        int x = q.top().n;        q.pop();        if (vis[x])continue;        vis[x] = 1;        for(int i = a[x];i;i = e[i].next){            int y = e[i].v;            if(e[i].w > l)                 e[i].ww = 1;            else                e[i].ww = 0;                            if (dis[x] + e[i].ww < dis[y]){                dis[y] = dis[x] + e[i].ww;                q.push((st){y,dis[y]});            }        }    }}bool check(int l) {    dijkstra(l);    if(dis[n] > k) return false;    else return true;}int main(){    scanf("%d %d %d",&n,&m,&k);    for(int i = 1;i <= m;i++){        int u,v,w;        scanf("%d %d %d",&u,&v,&w);        addedge(u,v,w);        addedge(v,u,w);        r = r < w ? w : r;    }    begin = 1;    end = n;    int l = 1;    while(l <= r) {        int mid = (l+r) >> 1;        if(check(mid)) {            ans = mid;            r = mid-1;        }        else             l = mid+1;    }    if(ans == 0) {        printf("-1");        return 0;    }    printf("%d\n",ans);}