「Bzoj 1614」「Usaco2007 Jan」Telephone Lines架设电话线 (二分+最短路)

BZOJ 1614
Luogu 1948
from: USACO 2008 Jan Sliver(USACO刷题第18题)

如果去掉一条路径上的边，肯定要优先删掉大的边，所以我们可以二分整条路径上的第$k+1$大的边，把比他小边的设为0，大的设为1
然后最后的$dis[n]$就是如果第$k+1$大的边为当前二分$mid$值，所需要免费的边数，即可判断可行性。
注意：无向图的边数组要开多两倍！！！牢记！！！

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define ms(i, j) memset(i, j, sizeof i)
using namespace std;

const int MAXN = 1000 + 5, MAXM = 10000 + 5, INF = 1000000000;

struct edge{int v, wi;}e[MAXM*2]; //无向边要*2！ 
int en, n, m, k;
vector<int> G[MAXN];

void ins(int u, int v, int wi) {
    en++, e[en].v = v, e[en].wi = wi, G[u].push_back(en);
    en++, e[en].v = u, e[en].wi = wi, G[v].push_back(en);    
}
void clean() {
    for (int i=0;i<=n;i++) G[i].clear();
    en = 0;
}
void init() {
    clean();
    for (int i=1;i<=m;i++) {
        int ai, bi, li; scanf("%d%d%d", &ai, &bi, &li);
        ins(ai, bi, li);
    }
}
int dis[MAXN], vi[MAXN];
bool check(int x) {
    queue<int> q;
    for (int i=1;i<=n;i++) dis[i] = INF, vi[i] = false;
    dis[1] = 0, q.push(1), vi[1] = true; 
    while (!q.empty()) {
        int u = q.front(); q.pop(); vi[u] = false;
        for (int i=0;i<G[u].size();i++) {
            edge p = e[G[u][i]];
            int v = p.v, w = p.wi;
            if (dis[v]>dis[u]+(p.wi>x)) {
                dis[v] = dis[u]+(p.wi>x);
                if (!vi[v]) q.push(v), vi[v] = true;
            }
        }
    }
    return dis[n]<=k;
}
void solve() {
    int ans = -1, l = 0, r = 1000005;
    while (l<r) {
        int mid = (l+r)>>1;
        if (check(mid)) {
            ans = mid;
            r = mid;
        } else l = mid + 1;
    }    
    printf("%d\n", ans);
}
int main() {
    #ifndef ONLINE_JUDGE
    freopen("1.in", "r", stdin);freopen("1.out", "w", stdout);
    #endif
    while (scanf("%d%d%d", &n, &m, &k)==3) init(), solve();
    return 0;
}