codeforce 379(div.2)
2016-11-19 00:37
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A、B略
C题 ——贪心，二分查找：
对于每一个a[i]， 在d中二分查找 s-b[i]，注意不要忘记计算速度为x时需要花费的最小时间，以及整数范围为64位整数
1 #include <cstdio>
2 #include <algorithm>
3 #include <cstring>
4
5 using namespace std;
6 const int maxn = 2*100000+10;
7 typedef long long LL;
8
9 int n, m, k;
10 int x, s;
11 int a[maxn], b[maxn], c[maxn], d[maxn];
12
13 int main() {
14 scanf("%d%d%d", &n, &m, &k);
15 scanf("%d%d", &x, &s);
16 a[0] = x;
17 b[0] = 0;
18 for(int i = 1; i <= m; i++) scanf("%d",&a[i]);
19 for(int i = 1; i <= m; i++) scanf("%d",&b[i]);
20 for(int i = 0; i < k; i++) scanf("%d",&c[i]);
21 for(int i = 0; i < k; i++) scanf("%d",&d[i]);
22 LL ans = LL(n)*x;
23 for(int i = 0; i <= m; i++) {
24 if(b[i] <= s) {
25 int temp = b[i];
26 int tt = upper_bound(d, d+k, s - temp) - d;
27 if(tt) {
28 tt--;
29 LL as = n - c[tt] > 0 ? LL(n - c[tt]) * a[i] : 0;
30 ans = min(ans, as);
31 }
32 else {
33 LL as = LL(n) * a[i];
34 ans = min(ans, as);
35 }
36 }
37 }
38 printf("%I64d\n", ans);
39 }
D 思路很简单，注意细节即可
1 #include <cstdio>
2 #include <vector>
3 #include <algorithm>
4 #include <cstring>
5 #include <cstdlib>
6 #include <cmath>
7 using namespace std;
8 const int maxn = 500000+10;
9
10 struct P{
11 char ch;
12 double dist;
13 P(char c, double d): ch(c), dist(d){}
14 bool operator < (const P& b)const {
15 return dist < b.dist;
16 }
17 };
18 vector<P> dig1, dig2, dig3, dig4;
19 vector<P> ver1, ver2, ver3, ver4;
20
21 int n;
22 int x0, y0_;
23 #define y0 y0_
24 void ins(char c, int dx, int dy) {
25 if(dx == dy) {
26 if(dx > 0) {
27 double dis = sqrt(double(dy)*dy + double(dx)*dx);
28 dig1.push_back(P(c, dis));
29 }
30 if(dx < 0) {
31 double dis = sqrt(double(dy)*dy + double(dx)*dx);
32 dig3.push_back(P(c, dis));
33
34 }
35 }
36 if(dx == -dy) {
37 if(dx > 0) {
38 double dis = sqrt(double(dy)*dy + double(dx)*dx);
39 dig4.push_back(P(c, dis));
40 }
41 if(dx < 0) {
42 double dis = sqrt(double(dy)*dy + double(dx)*dx);
43 dig2.push_back(P(c, dis));
44
45 }
46 }
47 if(dy == 0){
48 if(dx > 0) {
49 double dis = sqrt(double(dy)*dy + double(dx)*dx);
50 ver1.push_back(P(c,dis));
51 }
52 if(dx < 0) {
53 double dis = sqrt(double(dy)*dy + double(dx)*dx);
54 ver3.push_back(P(c,dis));
55
56 }
57 }
58 if(dx == 0){
59 if(dy > 0) {
60 double dis = sqrt(double(dy)*dy + double(dx)*dx);
61 ver2.push_back(P(c, dis));
62 }
63 if(dy < 0) {
64 double dis = sqrt(double(dy)*dy + double(dx)*dx);
65 ver4.push_back(P (c, dis));
66 }
67 }
68 }
69 int main () {
70 scanf("%d", &n);
71 scanf("%d%d", &x0, &y0);
72 for(int i = 0; i < n; i++) {
73 char ch;
74 int xx, yy;
75 getchar();
76 scanf("%c%d%d", &ch, &xx, &yy);
77 int dy = yy - y0, dx = xx - x0;
78 ins(ch , dx, dy);
79 }
80 sort(dig1.begin(), dig1.end());
81 sort(dig2.begin(), dig2.end());
82 sort(dig3.begin(), dig3.end());
83 sort(dig4.begin(), dig4.end());
84 sort(ver1.begin(), ver1.end());
85 sort(ver2.begin(), ver2.end());
86 sort(ver3.begin(), ver3.end());
87 sort(ver4.begin(), ver4.end());
88 if(dig1.size()) {
89 if(dig1[0].ch == 'Q' || dig1[0].ch == 'B') {
90 printf("YES\n");
91 return 0;
92 }
93 }
94 if(dig2.size()) {
95 if(dig2[0].ch == 'Q' || dig2[0].ch == 'B') {
96 printf("YES\n");
97 return 0;
98 }
99 }
100 if(dig3.size()) {
101 if(dig3[0].ch == 'Q' || dig3[0].ch == 'B') {
102 printf("YES\n");
103 return 0;
104 }
105 }
106 if(dig4.size()) {
107 if(dig4[0].ch == 'Q' || dig4[0].ch == 'B') {
108 printf("YES\n");
109 return 0;
110 }
111 }
112 if(ver1.size()) {
113 if(ver1[0].ch == 'Q' || ver1[0].ch == 'R') {
114 printf("YES\n");
115 return 0;
116 }
117 }
118
119 if(ver2.size()) {
120 if(ver2[0].ch == 'Q' || ver2[0].ch == 'R') {
121 printf("YES\n");
122 return 0;
123 }
124 }
125 if(ver3.size()) {
126 if(ver3[0].ch == 'Q' || ver3[0].ch == 'R') {
127 printf("YES\n");
128 return 0;
129 }
130 }
131 if(ver4.size()) {
132 if(ver4[0].ch == 'Q' || ver4[0].ch == 'R') {
133 printf("YES\n");
134 return 0;
135 }
136 }
137 printf("NO\n");
138 return 0;
139
140 }
E题：首先使用并查集将相同颜色的结点缩成一点，此时树中结点黑白相间，只需要求出树的直径（最长路）除以2就是答案。
1 #include <cstdio>
2 #include <cstring>
3 #include <vector>
4 #include <algorithm>
5 #include <queue>
6 using namespace std;
7
8 const int maxn = 200000+10;
9 vector<int> G[maxn];
10 void add(int v, int u) {
11 G[v].push_back(u);
12 G[u].push_back(v);
13 }
14 int n;
15 int col[maxn];
16 int col2[maxn];
17 int vis[maxn];
18 int pa[maxn];
19 int findpa(int x) {return x == pa[x] ? x : pa[x] = findpa(pa[x]);}
20 int from, ans;
21 void dfs(int u, int v, int d) {
22 if(d > ans) {
23 ans = d;
24 from = v;
25 }
26 for(int i = 0 ; i< G[v].size(); i++) {
27 int z = G[v][i];
28 if(z != u) {
29 dfs(v, z, d+1);
30 }
31 }
32 }
33 int U[maxn];
34 int V[maxn];
35 int main() {
36 scanf("%d", &n);
37 for(int i = 1; i <= n; i++)
38 scanf("%d", &col[i]);
39
40 for(int i = 1; i < n; i++)
41 scanf("%d%d", &U[i], &V[i]);
42 for(int i = 1; i <= n; i++) pa[i] = i;
43 for(int i = 1; i <= n; i++) {
44 int fu = findpa(U[i]);
45 int fv = findpa(V[i]);
46 if(col[fu] == col[fv]) {
47 pa[fu] = fv;
48 }
49 }
50 for(int i = 1; i <= n; i++) {
51 int fu = findpa(U[i]);
52 int fv = findpa(V[i]);
53 if(col[fu] != col[fv]) {
54 add(fu, fv);
55 }
56 }
57 ans = -1;
58 int u = pa[1];
59 for(int i = 0; i < G[u].size(); i++) {
60 int v = G[u][i];
61 dfs(u, v, 1);
62 }
63 u = from;
64 for(int i = 0; i < G[u].size(); i++) {
65 int v = G[u][i];
66 dfs(u, v, 1);
67 }
68 ans = (ans+1)/2;
69 printf("%d\n", ans);
70 }
F题：注意到 a&b + a|b = a+b
即可根据数组b， c的定义求出a，然后用求出的a反推数组b，c判断是否是解
详见代码：
1 #include <iostream>
2 #include <algorithm>
3 #include <cstring>
4
5 using namespace std;
6 const int maxn = 200000 + 10;
7 typedef long long LL;
8
9 LL a[maxn], b[maxn], c[maxn];
10 int num[64];
11 int n;
12 int main() {
13 ios::sync_with_stdio(false);
14 cin >> n;
15 for(int i = 0; i < n; i++) cin >> b[i];
16 for(int i = 0; i < n; i++) cin >> c[i];
17 LL sum = 0;
18 for(int i = 0; i < n; i++) {
19 a[i] = b[i] + c[i];
20 sum += a[i];
21 }
22 if(sum % (2*n)) {
23 cout << -1 << endl;
24 return 0;
25 }
26 sum /= 2*n;
27 for(int i = 0; i < n; i++) {
28 a[i] = (a[i] - sum);
29 if(a[i] % n) {
30 cout << -1 << endl;
31 return 0;
32 }
33 else{
34 a[i] /= n;
35 }
36 }
37 for(int i = 0; i < n; i++) {
38 LL temp = a[i];
39 for(int j = 0; j < 64; j++) {
40 num[j] += temp%2;
41 temp /= 2;
42 }
43 }
44 for(int i = 0; i < n; i++) {
45 LL b_ = 0, c_ = 0;
46 for(int j = 0; j < 64; j++) {
47 if(a[i] & (1LL << j)) {
48 b_ += num[j] * (1LL << j);
49 c_ += n * (1LL << j);
50 }
51 else c_ += num[j] * (1LL << j);
52 }
53 if(b_ != b[i] || c_ != c[i]) {
54 cout << -1 << endl;
55 return 0;
56 }
57 }
58 cout << a[0];
59 for(int i = 1; i < n; i++) {
60 cout << " " << a[i];
61 }
62 cout << endl;
63 return 0;
64 }
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