标签:
ID | name | status | one word | |
POJ 5437 | Alisha’s Party | 赛后AC、 | 优先队列,模拟。对时间t排序 |
#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <queue> using namespace std; struct Node { char name[210]; int val; int t; bool operator < (const Node &a) const { if (a.val != val) return a.val > val; else return a.t < t; } }node[150010]; struct Open { int t, per; }open[150010]; bool cmp(Open a, Open b) { return a.t < b.t; } Node rem[150010]; priority_queue<Node>que; int main() { int k, m, q; int t; scanf("%d", &t); while (t--) { memset(open, 0, sizeof(open)); memset(rem, 0, sizeof(rem)); scanf("%d%d%d", &k, &m, &q); for (int i=0; i<k; ++i) { scanf("%s%d", node[i].name, &node[i].val); node[i].t = i; } for (int i=0; i<m; ++i) { scanf("%d%d", &open[i].t, &open[i].per); } while (!que.empty()) { que.pop(); } sort(open, open+m, cmp); int cnt = 0; int t = 0; for (int i=0; i<m; ++i) { if (t < open[i].t) { for (int j=t; j < open[i].t; ++j) { que.push(node[j]); } t = open[i].t; } int num = 0; while(num < open[i].per && !que.empty() && cnt < k) { rem[cnt++] = que.top(); num++; que.pop(); } } if (open[m-1].t < k) { for (int i=open[m-1].t; i<k; ++i) { que.push(node[i]); } } while(!que.empty() && cnt < k) { rem[cnt++] = que.top(); que.pop(); } int a; for (int i=0; i<q; ++i) { scanf("%d", &a); if (i == 0) printf("%s", rem[a-1].name); else printf(" %s", rem[a-1].name); } printf("\n"); } return 0; }
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POJ 5438 | Ponds | 赛后AC | 循环链表 |
#include <queue> #include <stdio.h> #include <iostream> #include <algorithm> #include <string.h> using namespace std; #define maxn 100000+50 // 边的个数 #define maxm 10000+50 // 节点个数 int t, p, m, a, b; int val[maxm]; int head[maxm]; int deg[maxm]; bool vis[maxm]; long long ans, temp; int num; struct Edge { int v, next; }edge[maxn]; int totEdge; queue<int>que; void addEdge(int a, int b) { edge[totEdge].v = b; edge[totEdge].next = head[a]; head[a] = totEdge++; } void init() { memset(head, -1, sizeof(head)); memset(edge, 0, sizeof(edge)); memset(val, 0, sizeof(val)); memset(vis, 0, sizeof(vis)); memset(deg, 0, sizeof(deg)); totEdge = 0; ans = 0; while(!que.empty()) que.pop(); } void topSort() { for (int i=1; i<=p; ++i) { if (deg[i] == 0) vis[i] = true; if (deg[i] == 1) { vis[i] = true; que.push(i); } } while(!que.empty()) { int top = que.front(); //cout << top << "===\n"; que.pop(); for (int i=head[top]; i!=-1; i=edge[i].next) { int v = edge[i].v; //cout << v << "----\n"; deg[v]--; if (!vis[v]) { if (deg[v] == 0) vis[v] = true; else if (deg[v] == 1) { vis[v] = true; que.push(v); } } } //cout << "***\n"; } } void Dfs(int u, int fa) { // cout << val[u] << "===" << u << endl;; num += 1; temp += val[u]; vis[u] = true; for (int i=head[u]; i!=-1; i=edge[i].next) { int v = edge[i].v; if (!vis[v] && v != fa) { Dfs(v, u); // u开头的这条链上的最后一个子节点,如果是u ,就说明到环的开始了。 } } } int main() { cin >> t; while (t--) { init(); cin >> p >> m; for (int i=1; i<=p; ++i) { cin >> val[i]; } for (int i=0; i<m; ++i) { cin >> a >> b; addEdge(a, b); addEdge(b, a); deg[a]++; deg[b]++; } topSort(); for (int i=1; i<=p; ++i) { num = 0; temp = 0; if (!vis[i]) Dfs(i, 0); if (num & 1) { ans += temp; } } cout << ans << endl; } return 0; }
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POJ 5439 | Aggregated Counting | |||
POJ 5440 | Clock Adjusting | |||
POJ 5441 | Travel | 赛后AC | 并查集+离线化。注音可以休息、 |
// 先对边按权值排序,然后对询问按照权值排序。 // 每次在满足当前条件下,遍历。把两个端点所在的集合 // 并起来。此时增加的pair数就是2*num1*num2. // 因为已经对询问按照权值排序。所以后面的集合一定会覆盖前面的、 // 所以只要一直询问下去。记录就可以了。 // 有一点并查集离线查询的地方。 #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> using namespace std; #define maxn 201000 // 节点数 #define maxm 51000 // 询问数 #define tot 1001000 // 边数 int t, n, m, q; int num[maxn]; // 维护每个节点当前所在集合的节点个数、 int ans[maxm]; // 保存每次询问的结果、 int fa[maxn]; // 并查集 struct Node { int st, ed, dis; }node[tot]; int cmp(Node a, Node b) { return a.dis < b.dis; } struct Querry{ int id, ask; }que[maxm]; void init() { for (int i=1; i<=n; ++i) { fa[i] = i; num[i] = 1; } memset(ans, 0, sizeof(ans)); } int cmp2(Querry a, Querry b) { return a.ask < b.ask; } int find_(int i) { if (fa[i] == i) return i; else return fa[i] = find_(fa[i]); } int main() { int a, b, d; scanf("%d", &t); while (t--) { scanf("%d%d%d", &n, &m, &q); init(); for (int i=1; i<=m; ++i) { scanf("%d%d%d", &a, &b, &d); node[i].st = a; node[i].ed = b; node[i].dis = d; } sort(node+1, node+1+m, cmp); for (int i=0; i<q; ++i) { scanf("%d", &que[i].ask); que[i].id = i+1; } sort(que, que+q, cmp2); int now = 1; int tempans = 0; int flag = 0; for (int i=0; i<q; ++i) { flag = que[i].id; for(; node[now].dis<=que[i].ask && now<=m; now++) { int t1 = node[now].st; int t2 = node[now].ed; if (t1 == t2) { continue; } if (find_(t1) == find_(t2)) { continue; } else if (find_(t1) != find_(t2)) { tempans += 2*num[find_(t1)]*num[find_(t2)]; num[find_(t1)] += num[find_(t2)]; fa[find_(t2)] = find_(t1); } } ans[flag] += tempans; } for (int i=1; i<=q; ++i) { printf("%d\n", ans[i]); } } return 0; }
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POJ 5442 | Favorite Donut | |||
POJ 5443 | The Water Problem | 题如其名,水题。 | ..........就是水。 | |
POJ 5444 | Elven Postman | 赛后AC | 二叉树建和遍历。 |
#include <stdio.h> #include <string.h> #include <iostream> using namespace std; struct Tree { int l, r; }node[1010]; bool flag; void init() { memset(node, 0, sizeof(node)); } void addNode(int u, int c) { if (node[u].l==0 && u<c) {node[u].l = c;return;} else if (node[u].r==0 && u>c) {node[u].r = c;return;} else if (u<c) addNode(node[u].l, c); else addNode(node[u].r, c); } void find(int u, int c) { if (u == c) return; else if (u<c) { cout << ‘W‘; find(node[u].l, c); } else if (u > c) { cout << ‘E‘; find(node[u].r, c); } } int main() { int t, n, q, r, c; cin >> t; while(t--) { init(); cin >> n; cin >> r; node[r].l = node[r].r = 0; for (int i=1; i<n; ++i) { cin >> c; addNode(r, c); } cin >> q; for (int i=0; i<q; ++i) { flag = true; cin >> c; find(r, c); cout << endl; } } return 0; }
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POJ 5445 | Food Problem | |||
POJ 5446 | Unknown Treasure | |||
POJ 5447 | Good Numbers | |||
POJ 5448 | Marisa’s Cake | |||
POJ 5449 | Robot Dog |
很多题就是想不到思路。其实是可以做的。第一场只过了1道。。。。。。。。。人艰不拆。。。。。。
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原文地址:http://www.cnblogs.com/icode-girl/p/4818730.html