【玩耍】一水

【训练赛地址】点击打开链接

【PS】由于几乎是中文题目和题目比较短，就不说题意了。

【A】中文题目，状压+BFS，没有什么坑点，上代码。

【AC代码】

////Created by just_sort 2016/12/1//Copyright (c) 2016 just_sort.All Rights Reserved//#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tree_policy.hpp>#include <ext/pb_ds/hash_policy.hpp>#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;using namespace __gnu_pbds;typedef long long LL;typedef pair<int, LL> pp;#define MP(x,y) make_pair(x,y)const int maxn = 200005;typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>order_set;//headint n, m, tt, sx, sy;int dir[4][2] = {{0,1},{0,-1},{-1,0},{1,0}};char maze[22][22];bool vis[22][22][1<<10];struct node{    int x, y, t, state;    node(){}    node(int x, int y, int t, int state):x(x), y(y), t(t), state(state){}};int getid(char c){    if(islower(c)){        return c - 'a';    }    else{        return c - 'A';    }}int bfs(){    queue<node>q;    node now, nex;    now = node(sx, sy, 0, 0);    q.push(now);    vis[sx][sy][0] = 1;    while(!q.empty()){        now = q.front();        q.pop();        if(now.t >= tt) break;        for(int i = 0; i < 4; i++){            int dx = now.x + dir[i][0];            int dy = now.y + dir[i][1];            int t = now.t + 1;            int state = now.state;            if(vis[dx][dy][state] || dx < 0 || dx >= n || dy < 0 || dy >= m || maze[dx][dy] == '*'){                continue;            }            if(isupper(maze[dx][dy]) && !(state & (1<<getid(maze[dx][dy])))) continue;            if(islower(maze[dx][dy])) state |= (1<<getid(maze[dx][dy]));            if(maze[dx][dy] == '^') return t;            if(!vis[dx][dy][state]){                vis[dx][dy][state] = 1;                q.push(node(dx, dy, t, state));            }        }    }    return -1;}int main(){    while(scanf("%d%d%d",&n,&m,&tt)!=EOF)    {        tt -= 1;        memset(vis, false, sizeof(vis));        for(int i = 0; i < n; i++) scanf("%s", maze[i]);        for(int i = 0; i < n; i++){            for(int j = 0; j < m; j++){                if(maze[i][j] == '@'){                    sx = i, sy = j, maze[i][j] = '.';                    break;                }            }        }        int ans = bfs();        printf("%d\n",ans);    }    return 0;}

【B】先考虑每一行，再考虑一下每一行里面的元素，发现解决的这个问题有大量的重叠子问题，一个简单的dp想法就由然而生了。

【代码君】

////Created by just_sort 2016/12/1//Copyright (c) 2016 just_sort.All Rights Reserved//#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tree_policy.hpp>#include <ext/pb_ds/hash_policy.hpp>#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;using namespace __gnu_pbds;typedef long long LL;typedef pair<int, LL> pp;#define MP(x,y) make_pair(x,y)const int maxn = 200005;typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>order_set;//headint dp1[maxn], dp2[maxn];int n, m;int main(){    while(scanf("%d%d",&n,&m)!=EOF)    {        memset(dp1, 0, sizeof(dp1));        memset(dp2, 0, sizeof(dp2));        for(int i = 2; i <= n + 1; i++){            for(int j = 2; j <= m + 1; j++){                int x;                scanf("%d", &x);                dp1[j] = max(dp1[j-1], dp1[j-2]+x);            }            dp2[i] = max(dp2[i-1], dp2[i-2] + dp1[m + 1]);        }        printf("%d\n", dp2[n+1]);    }    return 0;}

【C】就是线段树的区间合并问题，不过这个题要维护的东西有点多，看代码吧。

////Created by just_sort 2016/12/1//Copyright (c) 2016 just_sort.All Rights Reserved////#include <ext/pb_ds/assoc_container.hpp>//#include <ext/pb_ds/tree_policy.hpp>//#include <ext/pb_ds/hash_policy.hpp>//#include <set>//#include <map>//#include <queue>//#include <stack>//#include <cmath>//#include <cstdio>//#include <cstdlib>//#include <cstring>//#include <iostream>//#include <algorithm>//using namespace std;//using namespace __gnu_pbds;//typedef long long LL;//typedef pair<int, LL> pp;//#define MP(x,y) make_pair(x,y)const int maxn = 200005;//typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>order_set;//head#include <cstdio>#include <iostream>#include <algorithm>using namespace std;struct node{    int l, r, len;    int l0, l1;//左端点开始的0,1的个数    int r0, r1;//右端点开始的0,1的个数    int max0, max1, cnt0, cnt1, flag;//区间连续0,1的个数，区间0,1的个数}Tree[maxn<<2];int n, m, A[maxn];void pushup(int rt){    Tree[rt].l0 = Tree[rt*2].l0;    if(Tree[rt*2].l0 == Tree[rt*2].len) Tree[rt].l0 += Tree[rt*2+1].l0;    Tree[rt].r0 = Tree[rt*2+1].r0;    if(Tree[rt*2+1].r0 == Tree[rt*2+1].len) Tree[rt].r0 += Tree[rt*2].r0;    Tree[rt].max0 = max(Tree[rt*2].max0, Tree[rt*2+1].max0);    Tree[rt].max0 = max(Tree[rt].max0, Tree[rt*2].r0 + Tree[rt*2+1].l0);    Tree[rt].cnt0 = Tree[rt*2].cnt0 + Tree[rt*2+1].cnt0;    Tree[rt].l1 = Tree[rt*2].l1;    if(Tree[rt*2].l1 == Tree[rt*2].len) Tree[rt].l1 += Tree[rt*2+1].l1;    Tree[rt].r1 = Tree[rt*2+1].r1;    if(Tree[rt*2+1].r1 == Tree[rt*2+1].len) Tree[rt].r1 += Tree[rt*2].r1;    Tree[rt].max1 = max(Tree[rt*2].max1, Tree[rt*2+1].max1);    Tree[rt].max1 = max(Tree[rt].max1, Tree[rt*2].r1 + Tree[rt*2+1].l1);    Tree[rt].cnt1 = Tree[rt*2].cnt1 + Tree[rt*2+1].cnt1;    //Tree[rt].len = Tree[rt*2].len + Tree[rt*2+1].len;}void Build(int l, int r, int rt){    Tree[rt].l = l, Tree[rt].r = r, Tree[rt].len = r - l + 1, Tree[rt].flag = -1;    if(l == r){        Tree[rt].l0 = Tree[rt].r0 = Tree[rt].max0 = Tree[rt].cnt0 = !A[l];        Tree[rt].l1 = Tree[rt].r1 = Tree[rt].max1 = Tree[rt].cnt1 = A[l];        return ;    }    int mid = (l + r) / 2;    Build(l, mid, rt*2);    Build(mid+1, r, rt*2+1);    pushup(rt);}void update(int L, int R, int rt, int cmd){    if(Tree[rt].l == L && Tree[rt].r == R){        if(cmd == 0)        {            Tree[rt].l0 = Tree[rt].r0 = Tree[rt].max0 = Tree[rt].cnt0 = Tree[rt].len;            Tree[rt].l1 = Tree[rt].r1 = Tree[rt].max1 = Tree[rt].cnt1 = 0;            Tree[rt].flag = 0;        }        else if(cmd == 1){            Tree[rt].l0 = Tree[rt].r0 = Tree[rt].max0 = Tree[rt].cnt0 = 0;            Tree[rt].l1 = Tree[rt].r1 = Tree[rt].max1 = Tree[rt].cnt1 = Tree[rt].len;            Tree[rt].flag = 1;        }        else if(cmd == 2){            swap(Tree[rt].l0, Tree[rt].l1);            swap(Tree[rt].r0, Tree[rt].r1);            swap(Tree[rt].max0, Tree[rt].max1);            swap(Tree[rt].cnt0, Tree[rt].cnt1);            Tree[rt].flag = 1 - Tree[rt].flag;        }        return ;    }    //lazy 操作    if(Tree[rt].flag >= 0){        update(Tree[rt*2].l, Tree[rt*2].r, rt*2, Tree[rt].flag);        update(Tree[rt*2+1].l, Tree[rt*2+1].r, rt*2+1, Tree[rt].flag);        Tree[rt].flag = -1;    }    int mid = (Tree[rt].l + Tree[rt].r) / 2;    if(R <= mid) update(L, R, rt*2, cmd);    else if(L > mid) update(L, R, rt*2+1, cmd);    else    {        update(L, mid, rt*2, cmd);        update(mid+1, R, rt*2+1, cmd);    }    pushup(rt);}node query(int L, int R, int rt){    if(L == Tree[rt].l && Tree[rt].r == R) return Tree[rt];    //lazy 操作    if(Tree[rt].flag >= 0){        update(Tree[rt*2].l, Tree[rt*2].r, rt*2, Tree[rt].flag);        update(Tree[rt*2+1].l, Tree[rt*2+1].r, rt*2+1, Tree[rt].flag);        Tree[rt].flag = -1;    }    int mid = (Tree[rt].l + Tree[rt].r) / 2;    if(R <= mid) return query(L, R, rt*2);    else if(L > mid) return query(L, R, rt*2+1);    else{        node t1, t2, t3;        t1 = query(L, mid, rt*2);        t2 = query(mid+1, R, rt*2+1);        t3.cnt1 = t1.cnt1 + t2.cnt1;        t3.l1 = t1.l1;        if(t1.l1 == t1.len) t3.l1 += t2.l1;        t3.r1 = t2.r1;        if(t2.r1 == t2.len) t3.r1 += t1.r1;        t3.max1 = max(t1.max1, t2.max1);        t3.max1 = max(t3.max1, t1.r1 + t2.l1);        t3.len = t1.len + t2.len;        return t3;    }}int main(){    int n, m;    while(scanf("%d%d",&n,&m) != EOF)    {        for(int i = 1; i <= n; i++) scanf("%d", &A[i]);        Build(1, n, 1);        int op, a, b;        for(int i = 1; i <= m; i++){            scanf("%d%d%d",&op,&a,&b);            a++;            b++;            if(op <= 2){                update(a, b, 1, op);            }            else{                node ans = query(a, b, 1);                if(op == 3){                    printf("%d\n",ans.cnt1);                }                else{                    printf("%d\n",ans.max1);                }            }        }    }    return 0;}

【D】贪心+排序，按比率排，避免小数，所以讲式子进行转换

【代码君】

////Created by just_sort 2016/12/1//Copyright (c) 2016 just_sort.All Rights Reserved//#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tree_policy.hpp>#include <ext/pb_ds/hash_policy.hpp>#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;using namespace __gnu_pbds;typedef long long LL;typedef pair<int, LL> pp;#define MP(x,y) make_pair(x,y)const int maxn = 200005;typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>order_set;//headstruct node{    int dps, hp;    node(){}    node(int dps, int hp):dps(dps), hp(hp){}    bool operator <(const node &rhs) const{        return (hp*rhs.dps) < (dps*rhs.hp);    }    void read(){        scanf("%d%d",&dps,&hp);    }}a[22];int  main(){    int n;    while(scanf("%d",&n)!=EOF)    {        for(int i = 0; i < n; i++) a[i].read();        sort(a, a + n);        int sum = 0;        int ans = 0;        for(int i = 0; i < n; i++){            sum = sum + a[i].dps;        }        for(int i = 0; i < n; i++){            ans = ans + sum * a[i].hp;            sum = sum - a[i].dps;        }        printf("%d\n", ans);    }    return 0;}

【E】 将不小于m的数看成1，剩下的看成0，只要区间1的个数不小于k就可以，枚举区间左端点，右端点可以通过two-pointers求出，时间复杂度O(n)

【代码君】

////Created by just_sort 2016/12/1//Copyright (c) 2016 just_sort.All Rights Reserved//#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tree_policy.hpp>#include <ext/pb_ds/hash_policy.hpp>#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;using namespace __gnu_pbds;typedef long long LL;typedef pair<int, LL> pp;#define MP(x,y) make_pair(x,y)const int maxn = 200005;typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>order_set;//headint a[maxn];int n, m, K;LL ans;int main(){    int T, n;    scanf("%d",&T);    while(T--){        scanf("%d%d%d", &n,&m,&K);        for(int i = 1; i <= n; i++) scanf("%d",&a[i]);        ans  = 0;        int r = 0, num = 0;        for(int i = 1; i <= n; i++){            while(num < K && r <= n){r++; num += (a[r] >= m);}            if(num < K) break;            ans += n - r + 1;            num -= (a[i] >= m);        }        printf("%lld\n",ans);    }    return 0;}