SDOI 2017 天才黑客

/*
根据claris的  博客以及 beginend 的博客来写的


首先考虑如何求出最短路  可以从样例看出  路径是从边走到边的， 所以我们将边看作点  有共同端点的两边之间
互相连边， 边权为lcp。 这条边自己的花费计算要拆点， 拆成的两个点之间连原来的花费  这样跑最短路就可以啦

然而这样的做法有问题， 考虑边数最大可能是M^2切要求M^2个lca  显然不行

这里采用claris巨佬的方法

/*********beginend的题解*******
假如现在有n个节点a[1..n]要两两求lca，我们将其按dfs序排序，设h[i]=dep[lca(a[i],a[i+1])]，根据后缀数组height数组的性质不难得到dep[lca(a[i],a[j])]=min(h[i]..h[j-1])。
那么我们可以枚举中间的每个h[i]，i两边的点就可以至少花费h[i]的费用来互相访问。我们只要建立前缀虚点和后缀虚电来优化连边即刻。

*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define M 100200
#define ll long long
#define id1(x) x * 2 - 1
#define id2(x) x * 2
using namespace std;
int n,m, k, deep[M], dfn[M], size[M], fa[M], son[M], top[M], ls[M], inh[M], outh[M], inx[M], outx[M], pst[M];
int  inp[M], outp[M], ins[M], outs[M], sz, head[M * 10], cnt, tm, to[M << 5], nxt[M << 5], ver[M << 5], a[M];
ll dis[M * 10];
bool vis[M * 10], in[M], out[M];
int read() {
    int nm = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for(; isdigit(c); c = getchar()) nm = nm * 10 + c -'0';
    return nm * f;
}

void init() {
    cnt = tm = 0;
    memset(head, 0, sizeof(head));
    memset(ls, 0, sizeof(ls));
    memset(inh, 0, sizeof(inh));
    memset(outh, 0, sizeof(outh));
    memset(son, 0, sizeof(son));
}

void push(int vi, int vj, int wei) {
    cnt++, to[cnt] = vj, ver[cnt] = wei, nxt[cnt] = head[vi], head[vi] = cnt;
    if(cnt >= (M << 5)) {
        puts("gg");
        exit(0);
    }
}

void add(int vi, int vj) {
    cnt++, to[cnt] = vj, nxt[cnt] = ls[vi], ls[vi] = cnt;
}

void dfs1(int now) {
    size[now] = 1;
    int maxx = 0;
    for(int i = ls[now]; i; i = nxt[i]) {
        int vj = to[i];
        deep[vj] = deep[now] + 1;
        fa[vj] = now;
        dfs1(vj);
        size[now] += size[vj];
        if(size[vj] > maxx) maxx = size[vj], son[now] = vj;
    }
}

void dfs2(int now) {
    dfn[now] = ++tm;
    if(son[now]) {
        top[son[now]] = top[now];
        dfs2(son[now]);
    }
    for(int i = ls[now]; i; i = nxt[i]) {
        int vj = to[i];
        if(vj == son[now]) continue;
        top[vj] = vj;
        dfs2(vj);
    }
}

void spfa() {
    for(int i = 1; i <= sz; i++) dis[i] = 10000000000000000ll;
    queue<int>q;
    q.push(0);
    dis[0] = 0;
    vis[0] = true;
    while(!q.empty()) {
        int op = q.front();
        q.pop();
        vis[op] = false;
        for(int i = head[op]; i; i = nxt[i]) {
            int vj = to[i];
            if(dis[vj] > dis[op] + ver[i]) {
                dis[vj] = dis[op] + ver[i];
                if(!vis[vj]) {
                    vis[vj] = true;
                    q.push(vj);
                }
            }
        }
    }
}

void dijk() {
    for(int i = 0; i <= sz; i++) dis[i] = 10000000000000000ll, vis[i] = 0;
    dis[0] = 0;
    priority_queue<pair<int,int> > q;
    q.push(make_pair(0,0));
    while(!q.empty()) {
        pair<int, int> now = q.top();
        q.pop();
        while(!q.empty() && vis[now.second]) now = q.top(), q.pop();
        if(vis[now.second]) break;
        int op = now.second;
        vis[op] = true;
        for(int i = head[op]; i; i = nxt[i]) {
            int vj = to[i];
            if(dis[vj] > dis[op] + ver[i]) {
                dis[vj] = dis[op] + ver[i];
                q.push(make_pair(-dis[vj],vj));
            }
        }
    }
}

int lca(int x, int y) {
    while(top[x] != top[y]) {
        if(deep[top[x]] < deep[top[y]]) swap(x, y);
        x = fa[top[x]];
    }
    return deep[x] < deep[y] ? x : y;
}

bool cmp(int x, int y) {
    return dfn[pst[x]] < dfn[pst[y]];
}

int main() {
    int T = read();
    while(T--) {
        init();
        n = read(), m = read(), k = read();
        sz = m * 2;
        for(int i = 1; i <= m; i++) {
            int vi = read(), vj = read(), wei = read();
            pst[i] = read();
            push(id1(i), id2(i), wei);
            inx[i] = inh[vj], inh[vj] = i, outx[i] = outh[vi], outh[vi] = i;
        }
        for(int i = 1; i < k; i++) {
            int vi = read(), vj = read();
            read();
            add(vi, vj);
        }
        dfs1(1), top[1] = fa[1] = 1;
        dfs2(1);
        for(int x = 1; x <= n; x++) {
            int a1 = 0;
            for(int i = inh[x]; i; i = inx[i]) a[++a1] = i, in[i] = 1;
            for(int i = outh[x]; i; i = outx[i]) a[++a1] = i, out[i] = 1;
            sort(a + 1, a + a1 + 1, cmp);
            for(int i = 1; i <= a1; i++) {
                inp[i] = ++sz;
                outp[i] = ++sz;
                if(in[a[i]]) push(id2(a[i]), inp[i], 0);
                if(out[a[i]]) push(outp[i], id1(a[i]), 0);
                if(i > 1) push(inp[i - 1], inp[i], 0), push(outp[i - 1], outp[i], 0);
            }
            for(int i = a1; i >= 1; i--) {
                ins[i] = ++sz;
                outs[i] = ++sz;
                if(in[a[i]]) push(id2(a[i]), ins[i], 0);
                if(out[a[i]]) push(outs[i], id1(a[i]), 0);
                if(i < a1) push(ins[i + 1], ins[i], 0), push(outs[i + 1], outs[i], 0);
            }
            for(int i = 1; i < a1; i++) {
                int lc = deep[lca(pst[a[i]], pst[a[i + 1]])];
                push(inp[i], outp[i + 1], lc);
                push(ins[i + 1], outs[i], lc);
            }
            for(int i = 1; i <= a1; i++) in[a[i]] = out[a[i]] = 0;
        }
        for(int i = outh[1]; i; i = outx[i]) push(0, id1(i), 0);
        //spfa();
        dijk();
        for(int x = 2; x <= n; x++) {
            ll mn = 10000000000000000ll;
            for(int i = inh[x]; i; i = inx[i]) mn = min(mn, dis[id2(i)]);
            cout << mn << "
";
        }
    }
    return 0;
}

/*
2
4 4 6
1 2 2 5
2 3 2 5
2 4 1 6
4 2 1 6
1 2 1
2 3 1
3 4 1
4 5 2
1 6 2

4 4 6
1 2 2 5
2 3 2 5
2 4 1 6
4 2 1 6
1 2 1
2 3 1
3 4 1
4 5 2
1 6 2

*/