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Codeforces Round #260 (Div. 2)

题目链接

A：水题，其实只要判断有没有一个ai != bi即可，因为都保证是1 - n的不相等数字

B：找到2 3 4的循环节，发现只有4和2，于是把大数%4，%2，在根据循环节去计算即可

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(i?1,0),dp(i?1,1)),
dp(i,1)=dp(i?1,0)+val[i]

D：Trie+博弈，根据字符串建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，如果能控制必胜又能控制必败就赢了，如果只能控制必胜，那么如果k为奇数也是赢，剩下都是输

E：并查集+贪心，对于每个集合，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表示集合最长长度即可

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\n");
    else printf("Poor Alex\n");
    return 0;
}

#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\n");
	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\n", ans);
    return 0;
}

#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\n", max(dp[100000][0], dp[100000][1]));
    return 0;
}

#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\n");
    else printf("Second\n");
    return 0;
}

#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\n", 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;
}

Codeforces Round #260 (Div. 2),布布扣,bubuko.com

Codeforces Round #260 (Div. 2)

标签：style   blog   http   color   os   io   for   2014   

原文地址：http://blog.csdn.net/accelerator_/article/details/38456219