本文整理汇总了C++中vi::resize方法的典型用法代码示例。如果您正苦于以下问题:C++ vi::resize方法的具体用法?C++ vi::resize怎么用?C++ vi::resize使用的例子?那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类vi
的用法示例。
在下文中一共展示了vi::resize方法的15个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的C++代码示例。
示例1: main
int main(){
while(scanf("%d",&N) == 1 && N){
int sum = 0,ans = INT_MAX,cmax = 0;
L.resize(N);
taken.resize(N);
loop(i,N) {
scanf("%d",&L[i]);
sum += L[i];
cmax = max(cmax,L[i]);
taken[i] = 0;
}
sort(all(L));
for(int i = sum;i>=1;i--){
if(sum%i == 0){
int l = sum/i,n = i;
if(l >= cmax){
fill(all(taken),0);
bool c = possibe(0,0,0,l,n);
if(c){
ans = l;
break;
}
}
}
}
printf("%d\n",ans);
}
示例2:
// Constructs the lookup table in O(V log V) time and memory
lowest_common_ancestor(const vvi& tree = {}, int root = 0) {
depth.resize(tree.size());
pos.resize(tree.size());
vector<visit> tour;
if (tree.size()) dfs(tree, tour, root, root);
euler = sparse_table<visit>(tour);
}
示例3: main
int main(){
freopen("1199.txt", "r", stdin);
//setbuf(stdout, NULL);
scanf("%d", &N);
adj = vvi(N, vi(N));
int total_edge = 0;
in_degree.resize(N); out_degree.resize(N);
for(int i = 0; i < N; i++){
for(int j = 0; j < N; j++){
scanf("%d", &adj[i][j]);
out_degree[i] += adj[i][j];
in_degree[j] += adj[i][j];
}
}
bool flag = false;
for(int i = 0; i < N; i++){
if( (out_degree[i] & 1) || (in_degree[i] & 1)){
flag = true;
break;
}
}
if(flag){
printf("-1\n");
}
else{
vi trace;
solve(0, trace);
for(int i = 0; i < trace.size(); i++){
printf("%d ", trace[i] + 1);
}
printf("\n");
}
}
示例4: lca
heavy_light(const vvi &graph, int root) {
links = lca(graph, 0);
st = seg_tree<T>(graph.size());
index.resize(graph.size()), jump.resize(graph.size());
dfs(root, root, root, graph);
}
示例5: main
int main(){
int N, M, a, b, c;
while(cin >> N >> M && N != 0 && M != 0){
AdjList.resize(N);
dfs_num.resize(N);
visited.resize(N);
dfs_low.resize(N);
numSCC = dfsNumberCounter = 0;
for(int i = 0; i < M; i++){
cin >> a >> b >> c;
AdjList[a-1].push_back(b-1);
if(c == 2) AdjList[b-1].push_back(a-1);
}
for(int i = 0; i < N; i++)
if(dfs_num[i] == 0)
tarjan(i);
if(numSCC == 1) cout << 1 << endl;
else cout << 0 << endl;
visited.clear();
AdjList.clear();
dfs_low.clear();
dfs_num.clear();
S.clear();
}
}
示例6: main
int main() {
int tc;
cin >> tc;
while (tc--) {
int n, m;
cin >> n >> m;
AdjList.clear(); AdjList.resize(n);
for (int i = 0; i < m; i++) {
int a, b;
cin >> a >> b;
AdjList[a].push_back(ii(b, 0));
}
bool cac = true;
int numScc = 0;
dfs_num.clear(); dfs_num.resize(n, UNVISITED);
dfs_parent.clear(); dfs_parent.resize(n, 0);
for(int i = 0; i < n; i++) {
if(dfs_num[i] == UNVISITED) {
numScc++;
if(graphCheck(i)) {
cac = false;
break;
}
}
}
if(cac && numScc == 1) cout << "YES\n";
else cout << "NO\n";
}
return 0;
}
示例7: init
void init() {
pertenece.resize(n);
rank.resize(n, 0);
size.resize(n, 1);
for(int i = 0; i < n; i++) {
pertenece[i] = i;
}
}
示例8: Task
Task(int _tl, int _testsCnt) {
tl = _tl;
testsCnt = _testsCnt;
averages.resize(testsCnt + 1, 0);
counters.resize(testsCnt + 1, 0);
sums.resize(testsCnt + 1, 0);
averages[testsCnt] = (tl / AVG_CONST) * testsCnt;
}
示例9: scc
void scc () {
nr.resize (gr.size (), -1);
low.resize (gr.size ());
inst.resize (gr.size (), false);
for (int i = 0; i < gr.size (); ++i) {
if (nr[i] < 0) {
visit (i);
}
}
}
示例10: init
void init(int n) {
h.resize(n, -1);
par.resize(n, -1);
//add 1
vi v1; v1.pb(1);
Node z1(1, 1, 0, v1);
G1.pb(z1);
h[1] = 1;
h[0] = 0;
}
示例11: main
int main () {
#ifndef ONLINE_JUDGE
freopen(INP, "r", stdin); freopen(OUT, "w", stdout);
#endif
scanf("%d", &n);
order.clear();
arr1.assign(n+1,0);
a1.clear();
FOR(i,1,n) scanf("%d", &a), a1.pb(a);
REP(i,n) arr1[a1[i]-1] = i+1;
REP(i,n) order[arr1[i]] = i;
while(scanf("%d", &a) != EOF){
arr.assign(n+1,0);
a2.clear();
a2.pb(a);
ans.resize(0);
len = 0;
FOR(i,1,n-1) scanf("%d", &a), a2.pb(a);
REP(i,n) arr[a2[i]-1] = i+1;
REP(i,n) updateLis(arr[i]);
cout << len << endl;
}
return 0;
}
示例12: clear
void clear()
{
dfs_num.clear();
dfs_num.resize(v+1);
connectedComp.clear();
cnt = 0;
}
示例13: stable_marriage
// calculates the "male optimal" stable marriage (i.e. there
// is no stable marriage where any man prefers his match
// over the one given here) a female optimal matching can be
// found by swapping men and women
// mpref: vector i is the list of i's preferred partners, in
// *decreasing* order of preference (0 indexed)
// fpref: as above, but for women
// match: a vector (passed in with any size) filled so that
// position i gives the man matched with woman i
void stable_marriage(const vvi& mpref, const vvi& fpref,
vi& match) {
// initially no one matched
match.resize(mpref.size(), -1);
// get a map from (w, m) to w's "rank" for m
vvi fmap(match.size(), vi(match.size()));
for(int i = 0; i < match.size(); ++i)
for(int j = 0; j < match.size(); ++j)
fmap[i][fpref[i][j]] = j;
vi next_prop(match.size(), 0);
queue<int> mfree;
for(int i = 0; i < match.size(); ++i)
mfree.push(i);
while(!mfree.empty()) {
const int m = mfree.front();
const int w = mpref[m][next_prop[m]];
mfree.pop();
if(match[w] == -1) {
match[w] = m;
}
else if(fmap[w][match[w]] > fmap[w][m]) {
mfree.push(match[w]);
match[w] = m;
}
else if(++next_prop[m] < match.size()) {
mfree.push(m);
}
}
}
示例14: main
int main()
{
ifstream fin("marathon.in");
ofstream fout("marathon.out");
int N;
fin >> N;
points.resize(N);
for (int i = 0; i < N; ++i)
{
fin >> points[i].first >> points[i].second;
}
T.resize(N - 2);
res.resize(N - 2);
T[0].first = 0;
T[0].second = dist_cont(2, N-1);
res[0] = T[0].first + T[0].second + dist_two(0, 2);
int result = res[0];
for (int k = 1; k <= N - 3; ++k)
{
T[k].first = T[k - 1].first + dist_two(k - 1, k);
T[k].second = T[k - 1].second - dist_two(k + 1, k+2);
res[k] = T[k].first + T[k].second + dist_two(k, k + 2);
result = min(result, res[k]);
}
fout << result << '\n';
return 0;
}
示例15: main
int main() {
ios::sync_with_stdio(0);
double d;
int m, tc = 1;
while( cin >> d >> m) {
cout << "Case " << tc++ << ":" << endl;
memo.clear(); memo.resize(105, -1);
prices.clear(); prices.resize(m + 1);
prices[0].first = 1; prices[0].second = (d + eps) * 100;
for(int i = 1; i <= m; i++) {
int tmp; cin >> tmp >> d;
prices[i].first = tmp; prices[i].second = (d + eps) * 100;
}
string str;
cin.get();
getline(cin, str);
stringstream ss(str);
int tmp;
while(ss >> tmp) {
cout << "Buy " << tmp << " for $" << fixed << setprecision(2) << (double)solve(tmp) / 100 << endl;
}
}
return 0;
}