2017-2018 ACM-ICPC Nordic Collegiate Programming Contest (NCPC 2017) 部分题／平衡树／最小环／思路bfs
交题地址
难度按照顺序递增
J - Judging Moose
队友敲的
#include B - Best Relay Team
队友敲的
#include
scanf("%lf%lf", &runers[i].f, &runers[i].s);
// }
double minimum = inf;
int res[5], cnt;
for(int i = 0; i < n; i++){
double temp = 0;
int tmplist[5];
cnt = 0;
temp += runers[i].f;
tmplist[cnt++] = i;
for(int j = 0; j < n; j++){
if(i != j){
temp += runers[j].s;
tmplist[cnt++] = j;
if(cnt == 4) break;
}
}
//cout << "%" << i <<" "<< temp <
if(temp - minimum < 0){
minimum = temp;
for(int j = 0; j < 4; j++) res[j] = tmplist[j];
}
}
printf("%.2lf\n",minimum);
for(int i = 0; i < 4; i++){
//cout <
cout << runers[res[i]].name << endl;
}
}
return 0;
} I - Import Spaghetti
在题目所给的有向图中找一个最小的环，并按照环的顺序逆向输出（任意起点皆可）
每个点bfs。。。居然在很智障的地方写re了（因为输入数组开小了哈哈哈）
#include int now;
if(q==root){
now = 1;
}else{
now = dis[q] + 1;
}
if(dis[mp[q][i]] > now){
que.push(mp[q][i]);
dis[mp[q][i]] = now;
}
}
}
return false;
}
int main(){
string name[maxn];
//freopen("29-496.4.in", "r", stdin);
int i,j,a,k,killme;
string str,now;
scanf("%d",&n);
for(i=1;i<=n;i++){
cin>>name[i];
change[name[i]] = i;
}
for(i=1;i<=n;i++){
cin>>str>>a;
int to = change[str];
for(j=1;j<=a;j++){
cin>>str;
getchar();
gets(in);
int p1 = 0;
int len = strlen(in);
for(k=0;kif(in[k] == ','){
in[k] = '\0';
now = in + p1;
mp[change[now]].push_back(to);
k += 2;
p1 = k;
continue;
}
}
now = in + p1;
mp[change[now]].push_back(to);
}
}
res = inf;
int pos = -1, maxx = inf;
for(i=1;i<=n;i++){
memset(vis, false, sizeof(vis));
bool flag = bfs(i);
if(res < maxx && flag){
maxx = res;
pos = i;
}
}
if(res == inf){
cout<<"SHIP IT\n";
return 0;
}
queue<int> que;
que.push(pos);
memset(dis, inf, sizeof(dis));
while(!que.empty()){
int q = que.front();
que.pop();
for(i=0;iint now;
if(q == pos){
now = 1;
}else{
now = dis[q] + 1;
}
if(dis[mp[q][i]] > now){
pre[mp[q][i]] = q;
que.push(mp[q][i]);
dis[mp[q][i]] = now;
}
if(mp[q][i] == pos){
pre[pos] = q;
goto here;
}
}
}
here:
int x = pos, p =0;
while(pre[x] != pos){
cout<" ";
x = pre[x];
}
cout<return 0;
} G - Galactic Collegiate Programming Contest
比赛时想会不会是平衡树，但是并不会写
套了wenwenla的平衡树板子一下就过了
附上wenwenla板子 github地址：https://github.com/wenwenla/ACM
#include >
struct Treap {
const static int NODECNT = maxn;
struct Node {
int ch[2], p, sz;
T v;
void make(int _l, int _r, int _p, const T& _v) {
ch[0] = _l; ch[1] = _r; p = _p; v = _v; sz = 1;
}
} node[NODECNT];
int m_rt, mp[NODECNT], mp_idx, node_idx;
void maintain(int x) {
node[x].sz = 1;
node[x].sz += node[x].ch[0] == -1 ? 0 : node[node[x].ch[0]].sz;
node[x].sz += node[x].ch[1] == -1 ? 0 : node[node[x].ch[1]].sz;
}
_Comp cmp;
explicit Treap(const _Comp& c) : cmp(c) { unsigned seed = 19971023; srand(seed); clear(); }
Treap() : cmp(_Comp()) { unsigned seed = 19971023; srand(seed); clear(); }
void clear() { m_rt = -1; mp_idx = -1; node_idx = 0; }
void ins(const T& val) { _ins(m_rt, val); }
void _ins(int& rt, const T& val) {
if(rt == -1) {
if(mp_idx == -1) { node[rt = node_idx++].make(-1, -1, rand(), val); }
else { node[rt = mp[mp_idx--]].make(-1, -1, rand(), val); }
} else {
int type = cmp(node[rt].v, val);
_ins(node[rt].ch[type], val);
maintain(rt);
if(node[rt].p < node[node[rt].ch[type]].p) rotate(rt, type);
}
}
void del(const T& val) { _del(m_rt, val); }
void _del(int& rt, const T& val) {
assert(rt != -1);
if(node[rt].v == val) {
if(node[rt].ch[0] == -1) {
mp[++mp_idx] = rt;
rt = node[rt].ch[1];
} else if(node[rt].ch[1] == -1) {
mp[++mp_idx] = rt;
rt = node[rt].ch[0];
} else {
int next = node[node[rt].ch[0]].p < node[node[rt].ch[1]].p;
rotate(rt, next);
_del(node[rt].ch[next ^ 1], val);
maintain(rt);
}
} else {
_del(node[rt].ch[cmp(node[rt].v, val)], val);
maintain(rt);
}
}
int find(const T& val) {
int rt = m_rt;
while(rt != -1) {
if(node[rt].v == val) return rt;
rt = node[rt].ch[cmp(node[rt].v, val)];
}
return -1;
}
void rotate(int& rt, int type) {
int tmp = node[rt].ch[type];
node[rt].ch[type] = node[tmp].ch[type ^ 1];
node[tmp].ch[type ^ 1] = rt;
maintain(rt); maintain(tmp);
rt = tmp;
}
int kth(int k) {
assert(k >= 1 && k <= size());
int rt = m_rt, res = -1;
while(rt != -1) {
int le = node[rt].ch[0] == -1 ? 0 : node[node[rt].ch[0]].sz;
if(le == k - 1) {
res = node[rt].v;
break;
} else if(le > k - 1) {
rt = node[rt].ch[0];
} else {
k -= le + 1;
rt = node[rt].ch[1];
}
}
return res;
}
int rank(const T& val) {
int rt = m_rt, cnt = 0;
while(rt != -1) {
int le = node[rt].ch[0] == -1 ? 0 : node[node[rt].ch[0]].sz;
if(cmp(node[rt].v, val)) {
cnt += le + 1;
rt = node[rt].ch[1];
} else {
rt = node[rt].ch[0];
}
}
return cnt + 1;
}
int size() { return node[m_rt].sz; }
};
Treapbool(*)(unit, unit)> tr (compare);
int main(){
tr.clear();
int n,m,i,j,a,b;
while(~scanf("%d%d",&n,&m)){
for(i=1;i<=n;i++){
save[i].prom = 0;
save[i].time = 0;
tr.ins(save[i]);
}
for(i=1;i<=m;i++){
scanf("%d%d",&a, &b);
tr.del(save[a]);
++save[a].prom;
save[a].time += b;
tr.ins(save[a]);
printf("%d\n",tr.rank(save[1]));
}
}
return 0;
} D - Distinctive Character
先讲一下设定，对于两个只含有0、1的串，它们的每一位如果相同则它们相似度+1
题目会给我们很多（1e5）这样的串，然后让我们给出一个01串，使得我们给出的串和题目输入的所有串的相似度最大值最小
炒鸡绕的，首先建图（最大1e6个点），把每个题目输入的串都当作起点，它们的相邻点是仅有一个位置和自身位置不同的串。bfs一遍，这些源串起点能到达的最远点即是问题所求。
#include if(in[j]-'0'){
save[i] += (1<0;
que.push(save[i]);
}
int farthest = -1,res = 0;
while(!que.empty()){
int q = que.front();
que.pop();
for(i=0;iint nnext = q ^ (1<if(dis[nnext] > dis[q]+1){
dis[nnext] = dis[q]+1;
que.push(nnext);
if(dis[nnext]>farthest){
res = nnext;
farthest = dis[nnext];
}
}
}
}
for(i=0;iif(res&(1<printf("1");
}else{
printf("0");
}
}
printf("\n");
}
return 0;
}