HDU 多校联合第二场

1001

不是在为这次比赛找借口。。。我真的想吐嘈一下这道题。。。

尼嘛！明明是状态dp！非得搞得数据随便贪心都能过！！！过你妹啊！埋头想dp的时候，大约开始十几分钟，刷新了一下。我了个去！快上百了！纯属恶心人吗？！ 吐嘈完毕 -_-!

 

状态dp：1 <= N <= 20，最多20位，完全可以位压缩。预处理一下hp_sum[i] (i = 010111...101)，表示i的二进制中所有为1的位置，他们的hp之和。 f[i] (i = 010111...101)表示i的二进制中所有为1的位置被干掉话费的最小代价

f[ i|(1<< j) ] = max(f[ i|(1<< j) ], f[i] + hp_sum[ i|(1<<j) ]*p[j].Dpi);

1002

对数据用数状数组预处理，当时根本就没有任何思路。。。任何思路。。。赛后各种水过。。。居然可以蹭数据。。。。枚举出来一个约数，去你妹的！

大概过程是对x，y进行离散话，然后按离散化以后的值作为位置，原值作为value加到数状数组里边。

统计到X时，比X小的数的和为sumx，个数为numx，所以时间为X*numx - sumx。同样，比mx大的数也能统计出来（总数减掉sumx）

同理对Y的在数状数组，也可以得到sumy, numy...

用到四颗数状数组：分别用来计算 sumx, numx, sumy, numy。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <ctime>
#include <queue>
#include <map>
#include <sstream>

#define CL(arr, val)    memset(arr, val, sizeof(arr))
#define REP(i, n)       for((i) = 0; (i) < (n); ++(i))
#define FOR(i, l, h)    for((i) = (l); (i) <= (h); ++(i))
#define FORD(i, h, l)   for((i) = (h); (i) >= (l); --(i))
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   x < y ? x : y
#define Max(x, y)   x < y ? y : x
#define E(x)    (1 << (x))
#define iabs(x)  ((x) > 0 ? (x) : -(x))

typedef long long LL;
const int eps = 1e-8;
const LL inf = 1e15;

using namespace std;

const int N = 100010;

struct node {
    int x;
    int y;
}p[N];

LL cx[N], cy[N];
LL Nx[N], Ny[N];
int tx[N], ty[N], n;


int lowbit(int i) {
    return i&(-i);
}
void add(LL c[], int i, LL val) {
    while(i <= n) {
        c[i] += val;
        i += lowbit(i);
    }
}

LL sum(LL c[], int i) {
    LL ret = 0;
    while(i > 0) {
        ret += c[i];
        i -= lowbit(i);
    }
    return ret;
}

int findx(int x) {
    int l = 1, r = n, mid;
    while(l <= r) {
        mid = (l + r) >> 1;
        if(tx[mid] == x)   return mid;
        else if(tx[mid] < x)   l = mid + 1;
        else    r = mid - 1;
    }
    return -1;
}

int findy(int x) {
    int l = 1, r = n, mid;
    while(l <= r) {
        mid = (l + r) >> 1;
        if(ty[mid] == x)   return mid;
        else if(ty[mid] < x)   l = mid + 1;
        else    r = mid - 1;
    }
    return -1;
}

void init() {
    CL(cx, 0); CL(cy, 0);
    CL(Nx, 0); CL(Ny, 0);
}

int main() {
    //freopen("data.in", "r", stdin);

    int t, i;

    int x, y;
    LL tolx, toly;
    LL sumx, sumy;
    LL numx, numy;
    LL tmpx, tmpy;
    LL ans;

    scanf("%d", &t);
    while(t--) {
        init();

        scanf("%d", &n);
        ans = inf;
        tolx = toly = 0;
        for(i = 1; i <= n; ++i) {
            scanf("%d%d", tx + i, ty + i);

            p[i].x = tx[i], p[i].y = ty[i];
            tolx += tx[i], toly += ty[i];
        }

        sort(tx + 1, tx + n + 1);
        sort(ty + 1, ty + n + 1);

        for(i = 1; i <= n; ++i) {
            x = findx(p[i].x);
            y = findy(p[i].y);

            add(cx, x, p[i].x); add(cy, y, p[i].y);
            add(Nx, x, 1);   add(Ny, y, 1);
        }
        //printf("tolx = %lld, toly = %lld\n",tolx,toly);
        for(i = 1; i <= n; ++i) {
            x = findx(p[i].x);
            y = findy(p[i].y);
            sumx = sum(cx, x) - p[i].x; sumy = sum(cy, y) - p[i].y;

            numx = sum(Nx, x) - 1; numy = sum(Ny, y) - 1;

            //printf("sumx = %lld, sumy = %lld, numx = %lld, numy = %lld\n", sumx, sumy, numx, numy);
            tmpx = (numx*p[i].x - sumx) + (tolx - sumx - p[i].x*(n - numx));
            tmpy = (numy*p[i].y - sumy) + (toly - sumy - p[i].y*(n - numy));

            //printf("tmpx(%lld) + tmpy(%lld) = %lld\n\n", tmpx, tmpy, tmpx + tmpy);
            ans = Min(ans, tmpx + tmpy);

        }
        cout << ans << endl;
    }
    return 0;
}

1003

各种蹭数据，真的好无聊。。。标程没看懂

找所有x的中点，所有y的中点，按到中点的距离排序。。。枚举前100个点。这都能蹭过去。。。。

1004

对于这题我只能说我是弱b。。。kruskal改动一下，按边权从大到小排序。。。。并查集维护，要删除的两个点不能在一个集合里边。。。

const int N = 100010;

int parent[N];

struct node {
    int x;
    int y;
    int d;
} p[N];

int n, k;
bool del[N];

bool cmp(node a, node b) {
    return a.d > b.d;
}

int find(int x) {
    int k = x, r = x, j;
    while(r != parent[r])   {
        if(del[r])  del[parent[r]] = true;
        r = parent[r];
    }

    while(k != r) {
        j = parent[k];
        parent[k] = r;
        k = j;
    }
    return r;
}

LL solve() {
    LL cnt = 0;
    int x, y;
    for(int i = 0; i < n - 1; ++i) {
        x = find(p[i].x);
        y = find(p[i].y);
        if(x == y)  continue;
        if(del[x] && del[y])   {cnt += LL(p[i].d); continue;}
        parent[y] = x;
        if(del[y])  del[x] = true;
        if(del[x])  del[y] = true;
    }
    return cnt;
}

void init() {
    CL(del, false);
    for(int i = 0; i <= n; ++i)     parent[i] = i;
}

int main() {
    //freopen("data.in", "r", stdin);

    int t, i, x;
    scanf("%d", &t);
    while(t--) {
        scanf("%d%d", &n, &k);
        init();
        for(i = 0; i < n - 1; ++i) {
            scanf("%d%d%d", &p[i].x, &p[i].y, &p[i].d);
        }
        for(i = 0; i < k; ++i) {
            scanf("%d", &x);
            del[x] = true;
        }
        sort(p, p + n - 1, cmp);
        cout << solve() << endl;
    }
    return 0;
}

  

    

1009

bfs 爆搞的，记录每个节点的最大剩余电量，防止有环的时候死循环

const int N = 2500010;
const int M = 50010;

struct node {
    int to;
    int next;
    double val;
} g[N];

int head[M], t;
double val[M];

void init() {
    CL(head, -1); t = 0;
}

void add(int u, int v, double w) {
    g[t].to = v; g[t].val = w; g[t].next = head[u];
    head[u] = t++;
}

void bfs(int s, double m) {
    val[s] = m;
    queue<int> q;
    q.push(s);

    int i, u, v;
    double w;

    while(!q.empty()) {
        u = q.front(); q.pop();
        for(i = head[u]; i != -1; i = g[i].next) {
            v = g[i].to;
            w = val[u]*(100.0 - g[i].val)/100.0;
            if(w > val[v]) {
                val[v] = w;
                q.push(v);
            }
        }
    }
}

int main() {
    //freopen("data.in", "r", stdin);

    int n, t, i, x;
    double y;
    while(~scanf("%d", &n)) {
        init();
        for(i = 1; i <= n; ++i) {
            scanf("%d", &t);
            while(t--) {
                scanf("%d%lf", &x, &y);
                add(i, x, y);
            }
            val[i] = -1;
        }
        scanf("%d%d%lf", &i, &x, &y);
        bfs(i, y);
        if(val[i] == -1)    puts("IMPOSSIBLE!");
        else    printf("%.2f\n", y - val[x]);
    }
    return 0;
}