• Codeforces Round #260 (Div. 2)


    Codeforces Round #260 (Div. 2)

    题目链接

    A:水题,事实上仅仅要推断有没有一个ai != bi就可以,由于都保证是1 - n的不相等数字

    B:找到2 3 4的循环节,发现仅仅有4和2,于是把大数%4,%2,在依据循环节去计算就可以

    C:dp,dp[i][0]表示不拿i数字。dp[i][1]表示拿i数字。状态转移为
    dp(i,0)=max(dp(i1,0),dp(i1,1)),
    dp(i,1)=dp(i1,0)+val[i]

    D:Trie+博弈。依据字符串建Trie,然后两遍dfs找出能控制自己必胜,和能控制自己必败的状态,假设能控制必胜又能控制必败就赢了,假设仅仅能控制必胜,那么假设k为奇数也是赢,剩下都是输

    E:并查集+贪心,对于每一个集合,两遍DFS能找出最长链,然后每次合并操作的时候,肯定是拿两遍的最长链中点去合并是最优的,这样带个权值len表示集合最长长度就可以

    代码:

    A:

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    
    const int N = 100005;
    int n, a[N], b[N];
    
    bool judge() {
        for (int i = 0; i < n; i++)
    	if (a[i] != b[i]) return true;
        return false;
    }
    
    int main() {
        scanf("%d", &n);
        for (int i = 0; i < n; i++)
    	scanf("%d%d", &a[i], &b[i]);
        if (judge()) printf("Happy Alex
    ");
        else printf("Poor Alex
    ");
        return 0;
    }

    B:

    #include <cstdio>
    #include <cstring>
    
    const int N = 100005;
    char str[N];
    
    int num[5][10];
    
    int main() {
        num[2][0] = 1; num[2][1] = 2; num[2][2] = 4; num[2][3] = 3;
        num[3][0] = 1; num[3][1] = 3; num[3][2] = 4; num[3][3] = 2;
        num[4][0] = 1; num[4][1] = 4;
        scanf("%s", str);
        if (strcmp(str, "0") == 0) {
    	printf("4
    ");
    	return 0;
        }
        int yu = 0;
        for (int i = 0; i < strlen(str); i++) {
    	yu = (yu * 10 + str[i] - '0') % 4;
        }
        int sb = yu % 2;
        int ans = (1 + num[2][yu] + num[3][yu] + num[4][sb]) % 5;
        printf("%d
    ", ans);
        return 0;
    }

    C:

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    
    const int N = 100005;
    int n;
    long long vis[N], dp[N][2];
    
    int main() {
        int a;
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) {
    	scanf("%d", &a);
    	vis[a]++;
        }
        for (long long i = 1; i <= 100000; i++) {
    	dp[i][0] = max(dp[i - 1][0], dp[i - 1][1]);
    	dp[i][1] = max(dp[i][1], dp[i - 1][0] + vis[i] * i);
        }
        printf("%lld
    ", max(dp[100000][0], dp[100000][1]));
        return 0;
    }

    D:

    #include <cstdio>
    #include <cstring>
    
    const int MAXNODE = 100005;
    const int SIGMA_SIZE = 26;
    
    struct Trie {
        int ch[MAXNODE][SIGMA_SIZE];
        int win[MAXNODE], lose[MAXNODE];
        int sz;
    
        void init() {
    	sz = 1;
    	memset(ch[0], 0, sizeof(ch[0]));
        }
    
        int idx(char c) {return c - 'a';}
    
        void insert(char *str) {
    	int n = strlen(str);
    	int u = 0;
    	for (int i = 0; i < n; i++) {
    	    int c = idx(str[i]);
    	    if (!ch[u][c]) {
    		memset(ch[sz], 0, sizeof(ch[sz]));
    		ch[u][c] = sz++;
    	    }
    	    u = ch[u][c];
    	}
        }
    
        int dfs1(int u) {
    	win[u] = 0;
    	for (int i = 0; i < SIGMA_SIZE; i++) {
    	    int v = ch[u][i];
    	    if (!v) continue;
    	    int tmp = dfs1(v);
    	    if (!tmp) win[u] = 1;
    	}
    	return win[u];
        }
    
        int dfs2(int u) {
    	lose[u] = 0;
    	int bo = 1;
    	for (int i = 0; i < SIGMA_SIZE; i++) {
    	    int v = ch[u][i];
    	    if (!v) continue;
    	    bo = 0;
    	    int tmp = dfs2(v);
    	    if (!tmp) lose[u] = 1;
    	}
    	if (bo) return lose[u] = 1;
    	return lose[u];
        }
    
        void getsg() {
    	dfs1(0);
    	dfs2(0);
        }
    };
    
    const int N = 100005;
    int n, k;
    char str[N];
    Trie gao;
    
    bool judge() {
        gao.init();
        scanf("%d%d", &n, &k);
        for (int i = 0; i < n; i++) {
    	scanf("%s", str);
    	gao.insert(str);
        }
        gao.getsg();
        if (gao.win[0] && gao.lose[0])
    	return true;
        if (gao.win[0]) {
    	if (k&1) return true;
    	return false;
        }
        return false;
    }
    
    int main() {
        if (judge()) printf("First
    ");
        else printf("Second
    ");
        return 0;
    }

    E:

    #include <cstdio>
    #include <cstring>
    #include <vector>
    using namespace std;
    
    const int N = 300005;
    int n, m, q, parent[N], len[N], vis[N], Maxu, Maxl;
    vector<int> g[N];
    
    void dfs(int u, int h, int p) {
        vis[u] = 1;
        int flag = 1;
        for (int i = 0; i < g[u].size(); i++) {
    	int v = g[u][i];
    	if (v == p) continue;
    	flag = 0;
    	dfs(v, h + 1, u);
        }
        if (flag) {
    	if (h > Maxl) {
    	    Maxl = h;
    	    Maxu = u; 
    	}
        }
    }
    
    int find(int x) {
        return x == parent[x] ? x : parent[x] = find(parent[x]);
    }
    
    int main() {
        scanf("%d%d%d", &n, &m, &q);
        for (int i = 1; i <= n; i++)
    	parent[i] = i;
        int u, v;
        while (m--) {
    	scanf("%d%d", &u, &v);
    	int pu = find(u);
    	int pv = find(v);
    	parent[pu] = pv;
    	g[u].push_back(v);
    	g[v].push_back(u);
        }
        for (int i = 1; i <= n; i++) {
    	if (vis[i]) continue;
    	Maxl = -1;
    	dfs(i, 0, 0);
    	Maxl = -1;
    	dfs(Maxu, 0, 0);
    	len[find(i)] = Maxl;
        }
        int c, a, b;
        while (q--) {
    	scanf("%d", &c);
    	if (c == 1) {
    	    scanf("%d", &a);
    	    printf("%d
    ", len[find(a)]);
    	}
    	else {
    	    scanf("%d%d", &a, &b);
    	    int pa = find(a);
    	    int pb = find(b);
    	    if (pa != pb) {
    		parent[pa] = pb;
    		len[pb] = max(len[pa], max(len[pb], (len[pb] + 1) / 2 + (len[pa] + 1) / 2 + 1));
    	    }
    	}
        }
        return 0;
    }


    版权声明:本文博主原创文章,博客,未经同意不得转载。

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  • 原文地址:https://www.cnblogs.com/zfyouxi/p/4906193.html
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