SRM 638 div2

2333...

Due to the small number of TC entries and the constant FST, I dropped to div2.

It's okay to go back to div1 ..

250

Question

500

Question ..

Just extend the BFS directly.

Pay attention to the heavy judgment. I also launched it with Kanto ..

1000

This question is understood as incorrect .. I told you why someone else's code looks wrong.

However, it is very easy to answer a question.

Which leaf nodes are burned by binary enumeration?

Then for each method

Minimum Short Circuit

Calculate the shortest path, enumeration side

Assume that the two nodes of the edge are u and the V weight is W.

The greatest (DIS [u] + dis [v] + W)/2 is the time when the data is burned out.

Why?

Assume that an edge is finally burned.

One is U, and V is burned by other nodes.

One is that u is burned by V.

First, set dis [u]> dis [v]

The answer is (L-(DIS [u]-Dis [v])/2 + dis [u] = (DIS [u] + dis [v] + l)/2

Second

Dis [v] + L = dis [u]

So the same dis [u] = (DIS [v] + L + dis [u])/2

Both of them can use this representation.

Then we will not divide it by 2 for convenience.

struct node {    int v, w;    node () {}    node (int _v, int _w) {v = _v; w = _w;}};vector<node>g[22];int ind[22], lea[22], pos[22], d[22], vis[22], q[1111];set<int> s;class CandleTimerEasy{public:    int differentTime(vector <int> A, vector <int> B, vector <int> len)    {        int n = A.size() + 1;        for(int i = 0; i < n; i++) g[i].clear();        memset(ind, 0, sizeof(ind));        for(int i = 0; i < n - 1; i++) {            g[A[i]].push_back(node(B[i], len[i]));            g[B[i]].push_back(node(A[i], len[i]));            ++ind[A[i]]; ++ind[B[i]];        }        s.clear();        int cnt = 0;        memset(pos, -1, sizeof(pos));        for(int i = 0; i < n; i++) {            if(ind[i] == 1) {                lea[cnt] = i;                pos[i] = cnt;                cnt++;            }        }        for(int sta = 1; sta < (1 << cnt); sta ++) {            int h = 0, t = 0;            for(int i = 0; i < n; i++) {                d[i] = INF; vis[i] = 0;                if(pos[i] != -1) {                    if(sta & (1 << pos[i])) {                        q[t++] = i;                        d[i] = 0;                        vis[i] = 1;                    }                }            }            while(h < t) {                int u = q[h++];                vis[u] = 0;                for(int i = 0; i < g[u].size(); i++) {                    int v = g[u][i].v;                    int w = g[u][i].w;                    if(d[u] + w < d[v]) {                        d[v] = d[u] + w;                        if(!vis[v]) {                            q[t++] = v;                            vis[v] = 1;                        }                    }                }            }            int mx = 0;            for(int i = 0; i < n - 1; i++) mx = max(mx, d[A[i]] + d[B[i]] + len[i]);            s.insert(mx);        }        return (int)s.size();    }}

SRM 638 div2