标签：put over define als pushd size ace ble clu
1.
Sample Input
6
-2 11 -4 13 -5 -2
10
-10 1 2 3 4 -5 -23 3 7 -21
6
5 -8 3 2 5 0
1
10
3
-1 -5 -2
3
-1 0 -2
0
Sample Output
20 11 13
10 1 4
10 3 5
10 10 10
0 -1 -2
0 0 0
//最基础的DP 数据结构作业
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100010;
int dp[MAXN]; //代表从第一个元素到i的子序列的最大
int a[MAXN];
int main () {
int K;
while(~scanf("%d", &K) && K) {
dp[0] = 0;
int s1 = -1, e1 = -1, s2 = - 1, e2 = -1;
int t = 0;
int tttt = -1;
memset(dp, 0, sizeof dp);
memset(a, 0, sizeof a);
for(int i = 1; i <= K; ++i) {
scanf("%d", &a[i]);
dp[i] = max(a[i], dp[i - 1] + a[i]);
if(a[i] < 0) t++;
if(a[i] == dp[i]) {
s1 = i; e1 = i;
}
else {
e1 = i;
}
if(dp[i] > tttt) {
e2 = e1;
s2 = s1;
tttt = dp[i];
}
}
if(t == K) printf("0 %d %d\n", a[1], a[K]);
else printf("%d %d %d\n", tttt, a[s2], a[e2]);
}
}
2.
#include <bits/stdc++.h>
using namespace std;
typedef pair<string, string> P;
map<string, string> MAP;
int main () {
MAP.insert(P("zero", "0"));
MAP.insert(P("one", "1"));
MAP.insert(P("two", "2"));
MAP.insert(P("three", "3"));
MAP.insert(P("four" ,"4"));
MAP.insert(P("five" ,"5"));
MAP.insert(P("six", "6"));
MAP.insert(P("seven", "7"));
MAP.insert(P("eight", "8"));
MAP.insert(P("nine", "9"));
string sb;
string t = "";
int a, b;
while(cin >> sb) {
if(sb == "=") {
//cout << t << endl;
b = std::stoi(t);
t = "";
if(a == 0 && b == 0) {
return 0;
}
cout << a + b << endl;
continue;
}
if(sb == "+") {
//cout << t << endl;
a = std::stoi(t);
t = "";
}
t += MAP[sb];
// t += (MAP.find(sb)->second);
// cout << (MAP.find(sb)->second) << endl;
}
}
3.
Sample Input
2
2
10 10
20 20
3
1 1
2 2
1000 1000
Sample Output
1414.2
oh!
//最小生成树 prim算法
#include <bits/stdc++.h>
using namespace std;
// const double MAXN = 0x3f3f3f3f;
#define MAXN 0xFFFFFFF
const int N = 101;
typedef pair<double, double> P;
std::vector<P> v;
double MAP[N][N];
bool vis[N];
int C; //小岛个数
double getD(P a, P b) {
return sqrt((a.first - b.first) * (a.first - b.first) + (a.second - b.second) * (a.second - b.second));
}
void buildMap() {
double l;
for(int i = 0; i < C; ++i) {
for(int j = i; j < C; ++j) {
l = getD(v[i], v[j]);
if(l >= 10 && l <= 1000) {
//cout << l << endl;
MAP[i][j] = MAP[j][i] = ((i == j) ? 0 : l);
}
else {
MAP[i][j] = MAP[j][i] = MAXN;
}
}
}
// for(int i = 0 ; i < C; ++i) {
// for(int j = 0; j < C; ++j) {
// cout << MAP[i][j] << " ";
// }
// cout << endl;
// }
}
void prim() {
int temp = C;
memset(vis, 0, sizeof vis);
vis[0] = 1;
double m;
double sum = 0;
int sb;
while(--temp) {
m = MAXN;
for(int i = 1; i < C; ++i) {
if(!vis[i] && MAP[0][i] < m){
//
m = MAP[0][i];
sb = i;
//cout << m << "???"<<endl;
}
}
if(m == MAXN) break;
vis[sb] = 1;
// cout << m << endl;
sum += m;
for(int i = 1; i < C; ++i) {
if(!vis[i] && MAP[sb][i] < MAP[0][i])
MAP[0][i] = MAP[sb][i];
}
}
//cout << temp << endl;
if(temp == 0) {
printf("%.1f\n", sum * 100.0);
}
else {
cout << "oh!" << endl;
}
}
int main () {
int T;
cin >> T;
while(T--) {
v.clear();
cin >> C;
for(int i = 0; i < C; ++i) {
double x, y;
cin >> x >> y;
v.push_back(P(x, y));
}
buildMap();
prim();
}
}
4.
Sample Input
3
1 2 1 0
1 3 2 0
2 3 4 0
3
1 2 1 0
1 3 2 0
2 3 4 1
3
1 2 1 0
1 3 2 1
2 3 4 1
0
Sample Output
3
1
0
//也是最小生成树 注意已经修好的路不用花钱
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 0x3fffffff;
const int N = 110;
bool vis[N];
int MAP[N][N];
int n;
int low[110];
void prim() {
memset(vis, 0, sizeof vis);
int ans = 0;
int k = 0;
vis[k] = 1;
// for(int i = 0; i < n; ++i) {
// if(!vis[i]) {
// low[i] = MAP[k][i];
// }
// }
int m;
for(int i = 1; i < n; ++i){
m = MAXN;
for(int i = 0; i < n; ++i){
if(!vis[i] && MAP[0][i] < m) {
k = i;
m = MAP[0][i];
}
}
ans += m;
vis[k] = 1;
for(int i = 0; i <= n + 1; ++i) {
if(!vis[i] && MAP[i][k] < MAP[0][i]) {
MAP[0][i] = MAP[i][k];
}
}
}
cout << ans << endl;
}
int main ()
{
ios::sync_with_stdio(false);
while(cin >> n) {
if(!n) break;
int x, y, s, b;
int t = (n - 1) * n / 2;
while(t--){
cin >> x >> y >> s >> b;
if(b) {
MAP[x-1][y-1] = MAP[y-1][x-1] = 0;
}
else {
MAP[x-1][y-1] = MAP[y-1][x-1] = s;
}
}
prim();
}
}
5.leetcode上的一道憨批题
//用O(n方）时间复杂度写出来的我着实憨批
class Solution {
public:
string pushDominoes(string dominoes) {
for(int i = 0; i < dominoes.size(); ++i) {
if(dominoes[i] != ‘.‘ && dominoes[i] != ‘M‘ && dominoes[i] != ‘S‘) {
continue;
}
int R, L;
R = 0x3f3f3f3f; L = 0x3f3f3f3f;
int s1, s2;
s1 = i; s2 = i;
char t1, t2;
for(int j = i + 1; j < dominoes.size(); ++j) {
if(dominoes[j] != ‘.‘ && dominoes[j] != ‘M‘ && dominoes[j] != ‘S‘) {
if(dominoes[j] == ‘L‘){
t1 = dominoes[j];
R = j - s1;
}
break;
}
}
if(i == 5)
cout << 66 << endl;
for(int j = i - 1; j >= 0; --j) {
if(j == 3 && i == 5) cout << dominoes[i] << endl;
if(dominoes[j] != ‘.‘ && dominoes[j] != ‘M‘ && dominoes[j] != ‘S‘) {
if(dominoes[j] == ‘R‘){
if(j == 3 && i == 5) cout << "sb" << endl;
t2 = dominoes[j];
L = s2 - j;
}
break;
}
}
if(i == 5) {
cout << R << " " << L << endl;
cout << dominoes[i] << endl;
}
if(R < L) {
dominoes[i] = t1 + 1;
}
else if(R > L) {
dominoes[i] = t2 + 1;
}
else {
dominoes[i] = ‘.‘;
}
}
for(int i = 0; i < dominoes.size(); ++i) {
if(dominoes[i] != ‘L‘ && dominoes[i] != ‘R‘ && dominoes[i] != ‘.‘) {
dominoes[i]--;
}
}
return dominoes;
}
};
//这是真的憨批题了
class Solution {
public:
bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {
if(rec1[2] > rec2[0] && rec1[0] < rec2[2] && rec1[3] > rec2[1] && rec1[1] < rec2[3])
return 1;
return 0;
}
};
//挺好的一道dp题
class Solution { public: int min(int a, int b) { return a > b ? b : a; } int numSquares(int n) { int dp[100000] = {0}; for(int i = 1; i <= n; ++i) { dp[i] = i; for(int j = 1; i - j * j >= 0; j++) { dp[i] = min(dp[i], dp[i - j * j] + 1); } } return dp[n]; } };
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标签：put over define als pushd size ace ble clu
原文地址：https://www.cnblogs.com/lightac/p/12521036.html
