Codeforces Round #260 (Div. 1) | 前3题

最近才重新做的，还是太挫。。

A：Boredom

类型：一个简单的dp（不过因为long long 的问题结果wa了好多发）

sum[i] ：表示选取当前 i 值；

sum2[i] ：表示不选当前 i 值；

转移方程：

sum[i] = sum2[i-1]+a[i]*i;

sum2[i] = max(sum[i-1], sum2[i-1]);

B：A Lot of Games

类型：dp+字典树

参考链接：http://www.cnblogs.com/LJ-blog/p/3904755.html

can_win数组：下一步由先手走，能否赢。能赢则为1，不能赢为0；

can_lose数组：下一步由先手走，能否输。能输则为1，不能输则为0；

应该从叶子节点往根推，可以肯定的是，叶子节点can_win为0，can_lose为1；

C：Civilization

类型：并查集+树的直径

一开始先根据所给的m条边建好图（邻接表），当然点与点的联系也要添加到并查集当中。

然后根据所建的图求出所有树的直径（再开一个数组保存，保存直径的数组的下标是该树并查集的根节点 ）。

查询：

操作1 直接find(x)寻找到x所在的树的根节点。

操作2 要让每次合并的树的直径尽量小，肯定是：树1的半径+树2的半径 + 1，树1的直径，树2的直径中的最大值。（半径往上取整）

AC代码：

A：

#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <queue>using namespace std;typedef long long ll;const int MAXN = 1e5 + 5;const int INF = 0x3f3f3f3f;int a[MAXN], n, mark[MAXN];ll sum[MAXN], sum2[MAXN];int main(){        while(scanf("%d", &n) != EOF)        {                int tmp;                memset(mark, 0, sizeof(mark));                memset(a, 0, sizeof(a));                memset(sum, 0, sizeof(sum));                memset(sum2, 0, sizeof(sum2));                int maxn = -1, minn = INF;                for(int i = 0;i < n; i++)                {                        scanf("%d", &tmp);                        a[tmp]++;                        maxn = max(maxn, tmp);                        minn = min(minn, tmp);                }                sum[minn] = (ll)a[minn]*minn;                for(int i = minn+1;i <= maxn; i++)                {                        if(a[i] == 0)                        {                            sum[i] = sum2[i] = max(sum[i-1],sum2[i-1]);                            continue;                        }                        sum[i] = sum2[i-1]+(ll)a[i]*i;                        sum2[i] = max(sum[i-1], sum2[i-1]);                        //cout<<"sum[i] = "<<sum[i]<<endl;                        //cout<<"sum2[i] = "<<sum2[i]<<endl;                }                cout<<max(sum[maxn], sum2[maxn])<<endl;        }        return 0;}

B：

#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>using namespace std;const int MAXN = 1e5+5;const int INF = 0x3f3f3f3f;const int SIGMA_SIZE = 30;typedef long long ll;char str[MAXN];int ch[MAXN][SIGMA_SIZE], sz;int can_lose[MAXN], can_win[MAXN];int insert(char *s){        int len = strlen(s);        int u = 0;        for(int i = 0;i < len; i++)        {                int c = s[i] - 'a';                if(!ch[u][c])  ch[u][c] = ++sz;                u = ch[u][c];        }}void dfs(int u){        bool hav = false;        for(int i = 0;i < 26; i++)        {                if(ch[u][i])                {                        dfs(ch[u][i]);                        hav = true;                }        }        if(!hav)        {                can_lose[u] = 1;                can_win[u] = 0;                return;        }        for(int i = 0;i < 26; i++)        {                int& v = ch[u][i];                if(v != 0 && can_lose[v] == 0)  can_lose[u] = 1;                if(v != 0 && can_win[v] == 0)   can_win[u] = 1;        }}int n, k;int main(){                scanf("%d%d", &n, &k);                sz = 0;                for(int i = 0;i < n; i++)                {                        scanf("%s", str);                        insert(str);                }                dfs(0);                //test                if((k%2 && can_win[0]&&!can_lose[0]) || (can_win[0]&&can_lose[0]))                        puts("First");                else                        puts("Second");        return 0;}

C：

#include <cstring>#include <cstdlib>#include <cstdio>#include <string>#include <vector>#include <iostream>using namespace std;const int MAXN = 6*1e5 + 5;const int INF = 0x3f3f3f3f;int n, m, f;int bcj[MAXN], val[MAXN];vector <int> edge[MAXN];bool vis[MAXN];void init(){        for(int i = 0;i <= n ; i++)                bcj[i] = i, edge[i].clear();        memset(val, 0, sizeof(val));        memset(vis, false, sizeof(vis));}int find(int u){        return bcj[u] == u ? u : bcj[u] = find(bcj[u]);}int tu;int dfs(int u, int len){        int val1 = 0, val2 = 0, tmp = 0, ret = len, noname;        vis[u] = true;        for(int i = 0;i < edge[u].size(); i++)        {                int v = edge[u][i];                if(vis[v])      continue;                noname = dfs(v, len+1);                tmp = noname - len;                ret = max(ret, noname);                if(tmp > val1) swap(val1, val2), val1 = tmp;                else if(tmp > val2)     val2 = tmp;                val[tu] = max(val[tu], val1+val2);        }        return ret;}int main(){        while(scanf("%d%d%d", &n, &m, &f) != EOF)        {                init();                int x, y;                for(int i = 0;i < m; i++)                {                        scanf("%d%d",&x, &y);                        edge[x].push_back(y);                        edge[y].push_back(x);                        x = find(x), y = find(y);                        bcj[x] = y;                }                //dfs                //                for(int i = 1;i <= n; i++)                {                        if(!vis[find(i)])       tu = find(i), dfs(tu, 0);                }                int co;                for(int i = 0;i < f; i++)                {                        scanf("%d", &co);                        if(co == 1)                        {                                scanf("%d", &x);                                //cout<<find(x)<<endl;                                printf("%d\n", val[find(x)]);                        }                        else                        {                                scanf("%d%d", &x, &y);                                x = find(x), y = find(y);                                if(x == y)      continue;                                int len1 = val[x]%2 == 0 ? val[x]/2 : val[x]/2+1;                                int len2 = val[y]%2 == 0 ? val[y]/2 : val[y]/2+1;                                val[x] = val[y] = max(max(val[x],val[y]), len1+len2+1);                                bcj[y] = x;                        }                }        }        return 0;}