今天多校不知道怎么说了
被坑不说，还这么弱。。。大家都过的题，我们没有过
后来还是一直补总算理解的还行
3979跟今天多校第一题，即4310一样
贪心就行
不知道DP怎么搞。。
View Code
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; struct node{ int hp, g; double a; friend bool operator < (const node &a, const node &b){ return a.a > b.a; } }p[10000 + 10]; int main(){ int tcase, z = 1; scanf("%d", &tcase); while(tcase --){ int n, m; scanf("%d%d", &n, &m); for(int i = 0; i < n; i ++){ scanf("%d%d", &p[i].hp, &p[i].g); p[i].a = 1.0 * p[i].g / ((p[i].hp - 1) /m +1); } sort(p, p + n); __int64 sum = 0; int t = 0; for(int i = 0; i < n; i ++){ t += (p[i].hp - 1) / m + 1; sum += t * p[i].g; //t++; } printf("Case #%d: %I64d\n", z++, sum); } return 0; }
4310就不多说了，暴力贪心的＝ =弱爆了
View Code
#include<cstdio> #include<algorithm> using namespace std; struct node{ int dps, hp; double sum; friend bool operator <(const node &a, const node &b) { return a.sum > b.sum; } }p[30]; int main() { int n; while(~scanf("%d", &n)) { for(int i = 0; i < n;i ++) scanf("%d%d", &p[i].dps, &p[i].hp), p[i].sum = ((double)p[i].dps) / p[i].hp; int ans = 0; int flag = 0; sort(p, p + n); for(int i = 0; i < n;i ++) { int tmp = 0; for(int j = i; j < n; j ++) { tmp += p[j].dps; } ans += (tmp * p[i].hp); //printf("--%d\n", ans); } printf("%d\n", ans); } return 0; }
4311最坑人了
一直把inf搞小了然后一直ＷＡ
这题数据弱
找中间值然后求附近100个点或者中位数之后暴力100个点都可以过，估计好多人是这样过的，只不过我们把inf取小了
标程的解法是我代码里的第一种
下面就引用吧
平面上两点间的 Manhattan 距离为 |x1-x2| + |y1-y2|
X 方向的距离与 Y 方向上的距离可以分开来处理。假设我们以 (xi,yi) 作为开会的地点，那么其余的点到该开会地点所需的时间为 X 方向上到 xi 所需要的时间加上 Y 方向上到 yi 所需要的时间。
对数据预处理后可以快速地求出x坐标小于xi的点的个数rankx, 并且这些 x 坐标值之和 sumx，那么这些点 X 方向上对结果的贡献为 rankx * xi - sumx；同理可以处理出 x 坐标大于 xi 的点在 X 方向上对结果的贡献值。同理可求得其余点在 Y 方向上到达 yi 所需要的总时间。
View Code
#include<cstdio> #include<cmath> #include<cstring> #include<algorithm> using namespace std; const int maxn = 100000 + 10; struct point{ __int64 x, y, sumx, sumy; __int64 sum; __int64 sumxx, sumyy; }p[maxn]; bool cmpx(point a, point b){ return a.x < b.x; } bool cmpy(point a, point b){ return a.y < b.y; } int main(){ int tcase; scanf("%d", &tcase); while(tcase --){ int n; scanf("%d", &n); for(int i = 1; i <= n; i ++){ scanf("%I64d%I64d", &p[i].x, &p[i].y); p[i].sumxx = p[i].sumyy = 0; } sort(p + 1, p + n + 1, cmpx); for(int i = 1; i <= n; i ++){ if(i == 1) p[i].sumx = p[i].x; else p[i].sumx = p[i - 1].sumx + p[i].x; } for(int i = 1; i <= n; i ++){ p[i].sumxx = i * p[i].x - p[i].sumx + p[n].sumx - p[i].sumx -(n - i ) * p[i].x; } sort(p + 1, p + n + 1, cmpy); for(int i = 1; i <= n; i ++){ if(i == 1) p[i].sumy = p[i].y; else p[i].sumy = p[i - 1].sumy + p[i].y; } for(int i = 1; i <= n; i ++){ p[i].sumyy = i * p[i].y - p[i].sumy + p[n].sumy - p[i].sumy - (n - i) * p[i].y; } __int64 ans = 1ll<<60; for(int i = 1; i <= n; i ++){ __int64 tmp = p[i].sumyy + p[i].sumxx; //printf("tmp: %d\n", tmp); if(tmp < ans) ans = tmp; } printf("%I64d\n", ans); } return 0; } /* struct point{ double x, y, dis; // friend bool operator < (const point &a, const point &b){ // return a.dis < b.dis; // } }p[maxn]; double distan(point a, point b){ return fabs(a.x - b.x) + fabs(a.y - b.y); } __int64 allDis(int n, point f){ __int64 sum = 0; for(int i = 0; i < n;i ++) sum += (__int64)distan(p[i], f); return sum; } bool cmp(point a, point b){ return a.dis < b.dis; } bool cmp1(point a, point b){ return a.x < b.x; } bool cmp2(point a, point b){ return a.y < b.y; } int main(){ int tcase; scanf("%d", &tcase); while(tcase --){ int n; scanf("%d", &n); point anss; //anss.x = anss.y = 0.0; for(int i = 0; i < n;i ++){ scanf("%lf%lf", &p[i].x, &p[i].y); //anss.x += p[i].x; //anss.y += p[i].y; } //anss.x /= n; //anss.y /= n; sort(p, p + n, cmp1); if(n & 1) anss.x = p[n / 2].x; else anss.x = (p[n/2].x + p[n/2-1].x)/2; sort(p, p + n, cmp2); if(n & 1) anss.y = p[n / 2].y; else anss.y = (p[n/2].y + p[n/2 -1].y)/2; for(int i = 0; i < n; i ++){ p[i].dis = distan(anss, p[i]); } sort(p , p + n, cmp); double tmp = ~0U>>1; int max_ = min(100, n); __int64 minn = (1ll<<60); for(int i = 0; i < max_; i ++){ __int64 temp = allDis(n, p[i]); if(minn > temp) minn = temp; } printf("%I64d\n", minn); } return 0; } struct point{ double x, y, dis; friend bool operator < (const point &a, const point &b){ return a.dis < b.dis; } }p[maxn]; double distan(point a, point b){ return fabs(a.x - b.x) + fabs(a.y - b.y); } double allDis(int n, point f){ double sum = 0; for(int i = 0; i < n;i ++) sum += distan(p[i], f); return sum; } bool cmp(point a, point b){ return a.dis < b.dis; } bool cmp1(point a, point b){ return a.x < b.x; } bool cmp2(point a, point b){ return a.y < b.y; } int main(){ int tcase; scanf("%d", &tcase); while(tcase --){ int n; scanf("%d", &n); point anss; //anss.x = anss.y = 0.0; for(int i = 0; i < n;i ++){ scanf("%lf%lf", &p[i].x, &p[i].y); //anss.x += p[i].x; //anss.y += p[i].y; } //anss.x /= n; //anss.y /= n; sort(p, p + n, cmp1); if(n & 1) anss.x = p[n / 2].x; else anss.x = (p[n/2].x + p[n/2-1].x)/2; sort(p, p + n, cmp2); if(n & 1) anss.y = p[n / 2].y; else anss.y = (p[n/2].y + p[n/2 -1].y)/2; for(int i = 0; i < n; i ++){ p[i].dis = distan(anss, p[i]); } sort(p , p + n, cmp); double tmp = (1ll<<60); int max_ = min(100, n); for(int i = 0; i < max_; i ++){ double temp = allDis(n, p[i]); if(tmp > temp) tmp = temp; } printf("%.0f\n",tmp); } return 0; } */
4313当时根本没想到是什么题
还是弱到爆啊
其实就是一个并查集
想到了没写。。也不敢写
估计一写又是一大堆BUG
其实就是看要不要加入那条边
View Code
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn = 100000 + 10; int pre[maxn]; struct node{ int u, v, w; friend bool operator < (const node &a, const node &b){ return a.w > b.w; } }p[maxn]; int find(int x){ int s; for(s = x; s != pre[s]; s = pre[s]); while(s != x){ int t = pre[x]; pre[x] = s; x = t; } return s; } int vis[maxn]; int main(){ int tcase; scanf("%d", &tcase); while(tcase --){ int n, k; scanf("%d%d", &n, &k); memset(pre, -1, sizeof(pre)); int u, v, w; for(int i = 0; i < n - 1; i ++){ scanf("%d%d%d", &u, &v, &w); p[i].u = u; p[i].v = v; p[i].w = w; pre[i] = i; } pre[n - 1] = n - 1; sort(p, p + n - 1); memset(vis, 0, sizeof(vis)); for(int i = 0; i < k; i ++) { scanf("%d", &u); vis[u] = 1; } __int64 sum = 0; int flag = 0; for(int i = 0; i < n - 1; i ++){ int u = find(p[i].u); int v = find(p[i].v); if(vis[u] && vis[v]) sum +=(__int64)p[i].w; else if(vis[u]) pre[v] = u; else if(vis[v]) pre[u] = v; else if(u < v) pre[v] = u; else pre[u] = v; } printf("%I64d\n", sum); } return 0; }
4318边改边过的，都不知道怎么回事就过了
其实就是个简单的BFS了
不过也可以用最短路
我觉得难写还是用了BFS了
View Code
#include<cstdio> #include<cstring> #include<algorithm> #include<queue> using namespace std; const double inf=0.0-(~0U>>1); const int maxn = 50000 + 10; struct edge { int to,w; edge*next; }*list[maxn]; void add_edge(int u,int v,int w) { edge *tmp=new edge; tmp->to=v; tmp->w=w; tmp->next=list[u]; list[u]=tmp; } struct point { int id; int sum; }; double p[maxn]; int vis[maxn]; void bfs(int start,int v) { queue<int>q; q.push(start); while(!q.empty()) { int hd = q.front(); q.pop(); double sum = p[hd]; edge *tmp = list[hd]; while(tmp != NULL) { /* if(vis[tmp->to] == 1) { tmp = tmp->next; continue; }*/ double cost = sum * (100.0 - tmp->w) / 100; if(cost > p[tmp->to]) { q.push(tmp->to); p[tmp->to] = cost; } tmp = tmp->next; /* p[tmp->to] = max(cost, p[tmp->to]); if(tmp->to != v) { q.push(tmp->to); vis[tmp->to] = 1; } tmp = tmp->next; //puts("asdf"); */ } } } int main() { int N; while(~scanf("%d", &N)) { memset(list, NULL, sizeof(list)); for(int i = 1; i <= N; i ++) { int k; scanf("%d", &k); while(k--) { int to, w; scanf("%d%d", &to, &w); add_edge(i, to, w); } p[i] = inf; } int u, v, sum; scanf("%d%d%d", &u, &v, &sum); //point start; //start.id = u; //start.sum = sum; //bfs(start); p[u] = sum; memset(vis, 0, sizeof(vis)); vis[u] = 1; bfs(u,v); if(p[v] == inf) puts("IMPOSSIBLE!"); else printf("%.2f\n", sum - p[v]); } return 0; }
唉。。。。等下看两集海贼王就去睡觉吧
今天也搞的累了
有时间把多校热身赛的线段树给搞了
不能再拖了
妹子，晚安