Codeforces Round #363 (Div. 2) A-C

A. Launch of Collider

找最近的R和L之间的距离

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN     freopen("input.txt","r",stdin);
#define FOUT    freopen("output.txt","w",stdout);
#define INF     0x3f3f3f3f
#define INFLL   0x3f3f3f3f3f3f3f
#define lson    l,m,rt<<1
#define rson    m+1,r,rt<<1|1
typedef long long LL;
typedef pair<int, int> PII;
using namespace std;

const int maxn = 200000 + 5;

char s[maxn];
int a[maxn];

int main() {
   //FIN
   int n;
   while(~scanf("%d", &n)) {
       scanf("%s", s + 1);
       int ans = INF;
       for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
       int flag = 0;
       for(int i = 1; i <= n; i++) {
           if(flag != 0) {
                if(s[i] == ‘R‘) flag = i;
                else if(s[i] == ‘L‘) ans = min(ans, a[i] - a[flag]);
           }
           else if(s[i] == ‘R‘) flag = i;
           else continue;
       }
       if(ans == INF) printf("-1\n");
       else printf("%d\n", ans/2);

   }
   return 0;
}

B. One Bomb

*是墙问可不可以用一个炸弹把所有的墙炸掉

炸弹可以炸当前坐标上x，y轴上的所有点 (就是一个十字

直接在读取数据时维护一个前缀然后再扫一遍判断就可以了

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN     freopen("input.txt","r",stdin);
#define FOUT    freopen("output.txt","w",stdout);
#define INF     0x3f3f3f3f
#define INFLL   0x3f3f3f3f3f3f3f
#define lson    l,m,rt<<1
#define rson    m+1,r,rt<<1|1
typedef long long LL;
typedef pair<int, int> PII;
using namespace std;

const int maxn = 1000 + 5;

int q1[maxn];
int q2[maxn];
char mp[maxn][maxn];

int main() {
   //FIN
   int n, m;
   while(~scanf("%d%d", &n, &m)) {
        int cnt = 0;
        memset(q1, 0, sizeof(q1));
        memset(q2, 0, sizeof(q2));
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                cin >> mp[i][j];
                if(mp[i][j] == ‘*‘) {
                    cnt++;
                    q1[i]++;
                    q2[j]++;
                }
            }
        }
        int flag = 0;

        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                int tmp = cnt;
                if(mp[i][j] == ‘*‘) tmp++;
                if(q1[i] + q2[j] == tmp) {
                    printf("YES\n");
                    printf("%d %d\n", i+1, j+1);
                    flag = 1;
                    break;
                }
            }
            if(flag) break;
        }
        if(flag == 0) printf("NO\n");

   }

   return 0;
}

C. Vacations

1 2 代表两种活动 3表示两种活动都可以进行 0是休息 不可以连续两天进行同一个活动，问最少休息的天数是多少， 贪心

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN     freopen("input.txt","r",stdin);
#define FOUT    freopen("output.txt","w",stdout);
#define INF     0x3f3f3f3f
#define INFLL   0x3f3f3f3f3f3f3f
#define lson    l,m,rt<<1
#define rson    m+1,r,rt<<1|1
typedef long long LL;
typedef pair<int, int> PII;
using namespace std;

const int maxn = 100 + 5;

int a[maxn];

int main() {
   //FIN
   int n;
   while(~scanf("%d", &n)) {
       for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
       int flag = -1;
       int ans = 0;
       for(int i = 1; i <= n; i++) {
            if(a[i] == 0) ans++, flag = -1;
            else if(a[i] == 1) {
                if(flag != 1) flag = 1;
                else flag = -1, ans++;
            }
            else if(a[i] == 2) {
                if(flag != 2) flag = 2;
                else flag = -1, ans++;
            }
            else {
                if(flag == 1) flag = 2;
                else if(flag == 2) flag = 1;
                //else if(a[i+1] == 1) flag = 2;
                //else flag = 1;
            }
       }
       printf("%d\n", ans);

   }
   return 0;
}