Codeforces Round 594 Div. 2
得出经验，打\(CF\)倒着开题比较爽。C题看了好久，没找到规律，看了看其他题，发现最后一题是个半裸的Tarjan
暴涨\(236\)分，飘了
不过新的名字颜色好丑，像\(\color{blue}{\text{电脑蓝屏的那种蓝}}\)
A. Integer Points
偶数会和偶数交于整数点，奇数会和奇数交于整数点
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;
LL read() {
LL k = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
k = k * 10 + c - 48, c = getchar();
return k * f;
}
int main() {
int t = read();
while(t--) {
LL odd1 = 0, odd2 = 0, even1 = 0, even2 = 0;
int n = read();
for(int i = 1; i <= n; ++i) {
int x = read();
if(x & 1) ++odd1;
else ++even1;
}
int m = read();
for(int i = 1; i <= m; ++i) {
int x = read();
if(x & 1) ++odd2;
else ++even2;
}
LL ans = odd1 * odd2 + even1 * even2;
printf("%lld\n", ans);
}
return 0;
}
B. Grow The Tree
如果没有限制，将这几根木棍拼成一条直线显然是最长的
但要求必须水平-竖直-水平-竖直……这样交错着拼
显然，使水平（或竖直）的木棍的长度之和尽量小，答案就会尽量大（因为此时拼出来的图形会趋近于一条直线）
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;
LL read() {
LL k = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
k = k * 10 + c - 48, c = getchar();
return k * f;
}
LL a[100010];
int main() {
int n = read();
LL sum1 = 0, sum2 = 0;
for(int i = 1; i <= n; ++i) a[i] = read(), sum2 += a[i];
sort(a+1, a+n+1);
for(int i = 1; i <= n/2; ++i) sum1 += a[i];
sum2 -= sum1;
sum1 *= sum1, sum2 *= sum2;
printf("%lld\n", sum1 + sum2);
return 0;
}
C. Ivan the Fool and the Probability Theory
找规律，并不会证明。一开始根本不会做，做完F题后回来找规律，最后压哨AC（最后这道题的分衰减到连B题都不如）
不妨设\(n < m\)
\(f_0 = 2,f_1=2, f_i = f_{i-1} + f_{i-2}, i \leq n\)
\(f2_1 =f_n, f2_i = f2_{i-1} + f_{i-2}, i \leq m\)
最后答案就是\(f2_m\)
翻了翻别人的代码，发现他们找的规律都是\(f_0 = 2,f_1=2, f_i = f_{i-1} + f_{i-2}, i \leq \max{n, m}\)，答案是\(f_n + f_m -2\)。其实都是一个道理 (只不过我的规律麻烦一些)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
#define mod 1000000007
using namespace std;
LL read() {
LL k = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
k = k * 10 + c - 48, c = getchar();
return k * f;
}
LL f[100010], f2[100010];
int main() {
int n = read(), m = read();
if(n > m) swap(n, m);
f[0] = f[1] = 2;
for(int i = 2; i <= m; ++i) f[i] = (f[i-1] + f[i-2]) % mod;
f2[1] = f[n];
for(int i = 2; i <= m; ++i) {
f2[i] = (f2[i-1] + f[i-2]) % mod;
}
cout << f2[m];
return 0;
}
F. Catowice City
说一说我的心路历程：
C题想了半天，没什么思路，先去看看别的题吧。D题：什么鬼题目，看不懂题意。E题：看不懂+1，而且感觉好麻烦，看看排行榜，woc，AC率好低。
心如死灰的打开F题：题意好歹能读懂，是个图论题，二分图染色？对着样例模拟了一下，不太对。又读了一遍题，画了画样例。等等，这不是道Tarjan缩点裸题吗。遂A之
排行榜上，我周围的人都A了 A、B、C、D1，我这个A了 A、B、F题的人显得如此的格格不入……
回归正题
将自环全部去掉，然后Tarjan缩点，如果缩成了一个点，则无解
否则，找一个入度为零的点，将它集合内的所有点标记为运动员，其他的点标记为陪审团
显然，这样构造一定是符合条件的
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
#define mod 1000000007
using namespace std;
LL read() {
LL k = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
k = k * 10 + c - 48, c = getchar();
return k * f;
}
struct zzz {
int f, t, nex;
}e[1000010 << 1]; int head[1000010], tot;
void add(int x, int y) {
e[++tot].t = y;
e[tot].f = x;
e[tot].nex = head[x];
head[x] = tot;
}
int vis[1000010], dfn[1000010], low[1000010], colnum, belong[1000010], deth, s[1000010], top;
int in[1000010];
void Tarjan(int x) {
low[x] = dfn[x] = ++deth;
s[++top] = x; vis[x] = 1;
for(int i = head[x]; i; i = e[i].nex) {
if(!dfn[e[i].t]) {
Tarjan(e[i].t);
low[x] = min(low[x], low[e[i].t]);
}
else
if(vis[e[i].t]) low[x] = min(low[x], dfn[e[i].t]);
}
if(dfn[x] == low[x]) {
++colnum; int k=0;
do {
k = s[top--];
belong[k] = colnum;
vis[k] = 0;
}while(k != x);
}
}
int main() {
int t = read();
while(t--) {
int n = read(), m = read(); tot = top = deth = colnum = 0;
for(int i = 1; i <= n; ++i)
head[i] = vis[i] = low[i] = belong[i] = dfn[i] = in[i] = 0;
for(int i = 1; i <= m; ++i) {
int x = read(), y = read(); if(x == y) continue;
add(x, y);
}
for(int i = 1; i <= n; ++i)
if(!dfn[i]) Tarjan(i);
if(colnum == 1) printf("NO\n");
else {
int cnt = tot, pos = 0;
for(int i = 1; i <= cnt; ++i)
if(belong[e[i].f] != belong[e[i].t])
++in[belong[e[i].t]];
for(int i = 1; i <= colnum; ++i)
if(!in[i]) pos = i;
int num = 0;
for(int i = 1; i <= n; ++i)
if(belong[i] == pos) ++num;
printf("YES\n%d %d\n", n - num, num);
for(int i = 1; i <= n; ++i)
if(belong[i] != pos) printf("%d ", i);
printf("\n");
for(int i = 1; i <= n; ++i)
if(belong[i] == pos) printf("%d ", i);
}
}
return 0;
}
