AtCoder ABC 183

本期考察STL的性质较多，如果熟悉的话基本可以秒掉

C - Travel

简单的搜索问题，练手

#define _CRT_SECURE_NO_WARNINGS

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LT(x) (x * 2)
#define RT(x) (x * 2 + 1)


using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;

int n, K;
int w[10][10];
bool vis[10];
int ans = 0;


void dfs(int step, int u, int val) {
    if (step == n - 1) {
        if (val + w[u][0] == K)
            ans += 1;
        return;
    }
    for (int v = 0; v < n; ++v) {
        if (vis[v] == 0) {
            vis[v] = 1;
            dfs(step + 1, v, val + w[u][v]);
            vis[v] = 0;
        }
    }
}


int main() { 
    //freopen("in.txt", "r", stdin);
    scanf("%d%d", &n, &K);
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            scanf("%d", &w[i][j]);
        }
    }
    vis[0] = 1;
    dfs(0, 0, 0);
    printf("%d\n", ans);
    return 0;
}

D - Water Heater

模拟，会排序就能做
用map来记录当前时间点的操作，每个时间点的操作完毕检查

#define _CRT_SECURE_NO_WARNINGS

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LT(x) (x * 2)
#define RT(x) (x * 2 + 1)


using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;

LL n, W;
map<LL, vector<pair<LL, LL>>> orders;


int main() { 
    //freopen("in.txt", "r", stdin);
    scanf("%lld%lld", &n, &W);
    set<LL> ts;
    for (int i = 0; i < n; ++i) {
        LL s, t, w;
        scanf("%lld%lld%lld", &s, &t, &w);
        orders[s].push_back({ 1, w });
        orders[t].push_back({ 2, w });
        ts.insert(s);
        ts.insert(t);
    }
    vi ti(ts.begin(), ts.end());
    sort(ti.begin(), ti.end());
    bool ac = 1;
    LL cur = 0;
    for (auto x : ti) {
        for (auto &item : orders[x]) {
            if (item.first == 1) {
                cur += item.second;
            }
            else {
                cur -= item.second;
            }
        }
        if (cur > W) {
            ac = 0;
            break;
        }
    }
    if (ac) {
        printf("Yes\n");
    }
    else {
        printf("No\n");
    }
    
    return 0;
}

E - Queen on Grid

由于可以走任意步，实际上就是在三个方向上求前缀和

#define _CRT_SECURE_NO_WARNINGS

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LT(x) (x * 2)
#define RT(x) (x * 2 + 1)


using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;

int H, W;
char s[2005][2005];
LL dp[2002][2002];
LL row_dp[2002][2002];      // sum
LL col_dp[2002][2002];
LL dia_dp[2002][2002];
LL mod = LL(1e9) + 7;


int main() { 
    //freopen("in.txt", "r", stdin);
    scanf("%d%d", &H, &W);
    for (int i = 0; i < H; ++i) {
        scanf("%s", s[i]);
    }
    dp[0][0] = 1;
    row_dp[0][0] = col_dp[0][0] = dia_dp[0][0] = 1;
    for (int i = 0; i < H; ++i) {
        for (int j = 0; j < W; ++j) {
            if (i + j == 0) continue;
            if (s[i][j] == '#') continue;
            if (i - 1 >= 0) dp[i][j] += col_dp[i - 1][j];
            if (j - 1 >= 0) dp[i][j] += row_dp[i][j - 1];
            if (i - 1 >= 0 && j - 1 >= 0) dp[i][j] += dia_dp[i - 1][j - 1];
            dp[i][j] %= mod;
            col_dp[i][j] = row_dp[i][j] = dia_dp[i][j] = dp[i][j];
            if (i - 1 >= 0)
                col_dp[i][j] = (col_dp[i - 1][j] + col_dp[i][j]) % mod;
            if (j - 1 >= 0)
                row_dp[i][j] = (row_dp[i][j - 1] + row_dp[i][j]) % mod;
            if (i - 1 >= 0 && j - 1 >= 0)
                dia_dp[i][j] = (dia_dp[i - 1][j - 1] + dia_dp[i][j]) % mod;
            
        }
    }
    LL ans = dp[H - 1][W - 1];
    printf("%lld\n", ans);
    return 0;
}

F - Confluence

并查集是并查集，但是在并查集里进行分类操作。乍一看是$O(N^2)$操作。实际上如果并到最后，并没有那么多的并操作。具体的时间复杂度待研究。。。。
用一个map[i][j]来记录第i个组去上j课的人数。并查集操作的时候小集加入大集。

#define _CRT_SECURE_NO_WARNINGS

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LT(x) (x * 2)
#define RT(x) (x * 2 + 1)


using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;

map<int, int> cnt[200020];
int n, q;
int c[200020];
int lev[200020];
int fa[200020];


int getf(int x) {
    if (x == fa[x]) return x;
    return fa[x] = getf(fa[x]);
}


int main() { 
    //freopen("in.txt", "r", stdin);
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= n; ++i) {
        scanf("%d", c + i);
        cnt[i][c[i]] = 1;
    }
    for (int i = 1; i <= n; ++i)
        fa[i] = i, lev[i] = 1;
    for (int i = 0; i < q; ++i) {
        int op, u, v;
        scanf("%d%d%d", &op, &u, &v);
        if (op == 1) {
            int fu = getf(u), fv = getf(v);
            if (fu == fv) continue;
            if (lev[fu] <= lev[fv]) {
                swap(fu, fv);
            }
            fa[fv] = fu;
            if (lev[fu] == lev[fv])
                lev[fu] += 1;
            for (auto it : cnt[fv]) {
                int ci = it.first;
                cnt[fu][ci] += it.second;
            }
        }
        else {
            int fu = getf(u);
            printf("%d\n", cnt[fu][v]);
        }
    }
    return 0;
}