本文整理汇总了C++中vii::pb方法的典型用法代码示例。如果您正苦于以下问题：C++ vii::pb方法的具体用法？C++ vii::pb怎么用？C++ vii::pb使用的例子？那么恭喜您, 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vii的用法示例。
在下文中一共展示了vii::pb方法的4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() {
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
scanf("%d", &n);
ll sum = 0;
for(int i = 0; i < n; ++i) {
int a, b; scanf("%d %d", &a, &b);
sum += a;
s.pb(mp(a, -1));
s.pb(mp(b, 1));
}
sort(s.begin(), s.end());
ll qtd = 0;
ll maxi = sum;
ll p = 0;
for(int i = 0; i < s.size(); ++i) {
if(s[i].S == -1) {
qtd -= s[i].S;
sum -= s[i].F;
}
ll foo = qtd * (ll)s[i].F + sum;
if(foo > maxi) {
maxi = foo;
p = s[i].F;
}
if(s[i].S == 1) qtd -= s[i].S;
}
printf("%lld %lld\n", p, maxi);
return 0;
}示例2: main
int main() {
int n,x,c;
scanf("%d", &n);
for(int i=0; i<n; i++){
scanf("%d %d", &x, &c);
m.pb(ii(x,c));
}
sort(m.begin(), m.end());
memset(dp, -1, sizeof dp);
printf("%d\n", m[0].second+recur(1,0));
return 0;
}示例3: main
int main(){
//maketestcase();
//freopen("input.txt", "r", stdin);
int n;
while(scanf("%d", &n) != EOF){
v.clear();
for(int i = 0; i < n; i++){
al[i].clear();
visited[i] = false;
}
for(int i = 0; i < n; i++){
int x, y;
scanf("%d %d", &x, &y);
v.pb(mkp(x,y));
}
map<ii, vi> m;
for(int i = 0; i < n; i++){
int x = v[i].first;
int y = v[i].second;
m[mkp(x/5, y/5)].pb(i);
}
int ix[5] = {0, 1, 1, 1, 0};
int iy[5] = {1, 1, 0, -1, -1};
repSTL(m, iter){
generateGraph(iter->second,iter->second);
ii p = iter->first;
for(int i = 0; i < 5; i++){
int x = p.first + ix[i];
int y = p.second + iy[i];
typeof(m.begin()) iter2 = m.find(mkp(x,y));
if(iter2 != m.end()){
generateGraph(iter->second, iter2->second);
}
}
}
m.clear();
int minS = 0;
for(int i = 0; i < n; i++){
if(!visited[i]){
ii p = bipartite(i);
minS += min(p.first, p.second);
}
}
printf("%d\n", minS);
}示例4: main
int main() {
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
scanf("%d %d", &k, &n);
for(int i = 0; i < n; ++i) {
scanf("%d", c + i);
foo.pb(mp(c[i], i));
}
sort(foo.begin(), foo.end());
for(int i = 0; i < n; ++i) {
int id = foo[i].S, x = foo[i].F;
int mini = min(x, k);
k -= mini;
sol[id] = mini;
}
for(int i = 0 ; i < n; ++i) {
if(i) printf(" ");
printf("%d", sol[i]);
}
printf("\n");
return 0;
}