[树链剖分+MST] LightOJ 1101 - A Secret Mission

LightOJ 1101 - A Secret Mission

题意：给一个n个点，m条边的无向连通图，每条边有一个权值，然后有q个询问，问从a到b的最小代价，一条路径的代价定义为这条路径上的最大的边权值。

题解：这个题一开始想了很久dijkstra，以为是某种应用或者可以快速更新，然而并没有什么结果。

后来就想暴力吧，想优化，于是先按边权值排序，然后加边，如果边两点已经连通了，那么就肯定没必要加这条边了，然后发现这不是kruskal吗？然后就变成一棵树了，然后就变成查询树上两点最大值了，然后就变成树链剖分模板题了...然后就AC了...代码几乎和上篇一样...

#include<bits/stdc++.h>using namespace std;const int N = 50005;struct eg{    int v, nex, w;    eg(){}    eg(int a, int b, int c){ v = a, w = b, nex = c; }}edg[N<<2];int fir[N], ecnt;void add(int a, int b, int c){    edg[ecnt] = eg(b, c, fir[a]);    fir[a] = ecnt++;    edg[ecnt] = eg(a, c, fir[b]);    fir[b] = ecnt++;}int fa[N], dep[N], siz[N], son[N], root;int tree[N<<2];void dfs(int rt){    siz[rt] = 1, son[rt] = 0;    for(int k = fir[rt]; k != -1; k = edg[k].nex){        if(edg[k].v != fa[rt]){            fa[edg[k].v] = rt;            dep[edg[k].v] = dep[rt] + 1;            dfs(edg[k].v);            if(siz[edg[k].v] > siz[son[rt]]) son[rt] = edg[k].v;            siz[rt] += siz[edg[k].v];        }    }}int w[N<<2], top[N],  cnt; //w[]为到线段树的映射void dfs2(int rt, int tp){    w[rt] = ++cnt; top[rt] = tp;    if(son[rt]) dfs2(son[rt], top[rt]);    for(int k = fir[rt]; k != -1; k = edg[k].nex){        if(edg[k].v != son[rt] && edg[k].v != fa[rt]){             dfs2(edg[k].v, edg[k].v);        }    }}void update(int rt, int l, int r, int pos, int val){    if(l == r){        tree[rt] = val;        return;    }    int mid = (l+r) >> 1;    if(pos <= mid) update(rt<<1, l, mid, pos, val);    else update(rt<<1|1, mid+1, r, pos, val);    tree[rt] = max(tree[rt<<1], tree[rt<<1|1]);}int n, m;struct E{    int u, v, w;    bool operator < (E a) const{        return w < a.w;    }}sav[N<<1];int seed[N];int find(int x){ return seed[x] < 0? x : seed[x] = find(seed[x]); }int join(int a, int b){    a = find(a), b = find(b);    if(a == b) return 0;    if(seed[a] > seed[b]) seed[a] = b;    else seed[b] = a;    return 1;}bool in[N<<1];void init(){    scanf("%d %d", &n, &m);    memset(fir, -1, sizeof(fir));    memset(siz, 0, sizeof(siz));    memset(tree, 0, sizeof(tree));    memset(seed, -1, sizeof(seed));    root = (n+1)/2;    fa[root] = cnt = ecnt = dep[root] = 0;    int a, b, c;    for(int i = 0; i < m; ++i) scanf("%d%d%d", &sav[i].u, &sav[i].v, &sav[i].w), in[i] = 0;    sort(sav, sav+m);    for(int i = 0; i < m; ++i){        if(join(sav[i].u, sav[i].v)) add(sav[i].u, sav[i].v, sav[i].w), in[i] = 1;    }    dfs(root);    dfs2(root, root);    for(int i = 0; i < m; ++i){        if(!in[i]) continue;        if(dep[sav[i].u] > dep[sav[i].v]) swap(sav[i].u, sav[i].v);        update(1, 1, cnt, w[sav[i].v], sav[i].w);    }}int maxi(int rt, int l, int r, int ql, int qr){    if( ql > r || qr < l) return 0;    if(ql <= l && qr >= r) return tree[rt];    int mid = (l+r) >> 1;    return max(maxi(rt<<1, l, mid, ql, qr), maxi(rt<<1|1, mid+1, r, ql, qr));}int solve(int va, int vb){    int f1 = top[va], f2 = top[vb], tmp = 0;    while(f1 != f2){        if(dep[f1] < dep[f2]){            swap(f1, f2);            swap(va, vb);        }        tmp = max(tmp, maxi(1, 1, cnt, w[f1], w[va]));        va = fa[f1], f1 = top[va];    }    if(va == vb) return tmp;    if(dep[va] > dep[vb]) swap(va, vb);    return max(tmp, maxi(1, 1, cnt, w[son[va]], w[vb]));}int main(){    int T, ca = 1;    scanf("%d", &T);    while(T--){        init();        int q;        scanf("%d", &q);        printf("Case %d:\n", ca++);        while(q--){            int a, b;            scanf("%d%d", &a, &b);            printf("%d\n", solve(a, b));        }    }}