Codeforces Round #234 (Div. 2)

Problems # Name A Inna and Choose Options standard input/output 1 s, 256 MB x1942 B Inna and New Matrix of Candies standard input/output 1 s, 256 MB x1556 C Inna and Huge Candy Matrix standard input/output 2 s, 256 MB x1114 D Dima and Bact

Problems

#	Name
A	Inna and Choose Options standard input/output 1 s, 256 MB		x1942
B	Inna and New Matrix of Candies standard input/output 1 s, 256 MB		x1556
C	Inna and Huge Candy Matrix standard input/output 2 s, 256 MB		x1114
D	Dima and Bacteria standard input/output 2 s, 256 MB		x371
E	Inna and Binary Logic standard input/output 3 s, 256 MB		x169

A题：直接暴力枚举每种情况即可。水题

B题：记录下S，G的距离，每种距离只要一次，开个vis数组标记，最后遍历一遍看又几个即可

C题：模拟旋转即可。

D题：并查集+floyd，把w=0的边的点并查集处理，判断同种类是否都在一个集合内，不是就No，剩下的就利用floyd求出最短路即可。

E题：位运算，每个数字对应的每个位向左和向右延生，这个区间内向下的那个三角形区间一定是会增加(1

代码：

A题：

#include 
#include 

int t, n;
char str[15];
char save[15][15];

bool judge(int a, int b) {
    int i, j;
    for (i = 0; i 

B题：
#include 
#include 
#include 
using namespace std;
const int N = 1005;
int n, m, i, j, vis[N];
char g[N][N];

int main() {
    scanf("%d%d", &n, &m);
    for (i = 0; i 

C题：
#include 
#include 
#include 
using namespace std;
int n, m, x, y, z, p, i, j;
struct Point {
    int x, y;
} po[100005];

void at(Point &a) {
    int x = a.x, y = a.y;
    a.y = n - x + 1;
    a.x = y;
}

void ht(Point &a) {
    int x = a.x, y = a.y;
    a.y = m - y + 1;
    a.x = x;
}

void ct(Point &a) {
    int x = a.x, y = a.y;
    a.y = x;
    a.x = m - y + 1;
}

int main() {
    scanf("%d%d%d%d%d%d", &n, &m, &x, &y, &z, &p);
    x %= 4;
    y %= 2;
    z %= 4;
    for (i = 0; i 

D题：
#include 
#include 
#include 
#define INF 0x3f3f3f3f
#define min(a,b) ((a) w) {
            f[type[u]][type[v]] = w;
            f[type[v]][type[u]] = w;
        }
        if (w == 0) {
            int pu = find(u);
            int pv = find(v);
            if (pu != pv)
                fa[pv] = pu;
        }
    }
    for (i = 2; i 

E题：
#include 
#include 

const int N = 100005;
const int M = 20;
int n, m, i, j, b;
int a[N][M];
__int64 sum, mi[32];

int main() {
    mi[0] = 1;
    for (i = 1; i