SDUACM暑期集训周赛（二）

POJ 3613 Cow Relays

[Problem]

给定m(<=100)条边的无向图，询问从s点到t点经过k条边的最短路径。 k (1e6)

[Solution]

我们定义一种操作（类似于矩阵乘）
c[i][j] = min(a[i][k] + b[k][j]) (1 <=k<=n )
容易证明此操作满足结合律
令该图的矩阵为a，则答案为a^k

[Code]

#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
#define maxn 200
#define inf 0x3f3f3f3f3f3f3f3f
int n, m, cnt, s, t, tim;
map<int , int > conn;
int id(int x)
{
    if (!conn.count(x)) conn[x] = ++cnt;
    return conn[x];
}
struct matrix {
   ll m[maxn][maxn];
   matrix(){
      memset(m, inf, sizeof(m));
   }
};
matrix mul(matrix a, matrix b)
{
    matrix c;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++){
            c.m[i][j] = inf;
            for(int k = 1; k <= n; k++)
                c.m[i][j] = min(c.m[i][j], a.m[i][k] + b.m[k][j]);
        }
    }
    return c;
}
matrix quickpow(matrix a, int b)
{
    matrix tot = a;
    b--;
    while(b > 0){
        if (b % 2 == 1) tot = mul(tot, a);
        a = mul(a, a);
        b /= 2;
    }
    return tot;
}
int main()
{
   // freopen("b.in", "r", stdin);
    scanf("%d%d%d%d", &tim, &m, &s, &t);
    cnt = 0;
    matrix base;
    for(int i = 1; i <= m; i++){
        int x, y, z;
        scanf("%d%d%d", &z, &x, &y);
        x = id(x);  y = id(y);
        base.m[x][y] = z;
        base.m[y][x] = z;
    }
    n = cnt;
    s = id(s); t = id(t);
    ll ans = quickpow(base, tim).m[s][t];
    cout << ans;
    return 0;
}

POJ 3190 Stall Reservations

[Solution]

首先把每头牛的区间排序，按照升序依次为每头牛分配stall
把每个stall的结束时间放到一个优先队列，每次取出结束时间最靠前的机器，判断这个stall是否可以为这头牛服务，否的话加入一个新的stall

[Code]

#include
#include
#include
#include
using namespace std;
typedef long long ll;
#define N 220000
int n, m;
struct node{
    int pos, l, r;
    bool operator <(const node &a) const {
        if (r == a.r)  return l > a.l;
        return r > a.r;
    }
} a[N];
int ans[N];
priority_queue q;
bool cmp(node a, node b)
{
    if (a.l == b.l)  return a.r < b.r;
    return a.l < b.l;
}
int main()
{
    //freopen("b.in", "r", stdin);
  while(~scanf("%d", &n)){
    int cnt = 0;
    while(!q.empty())  q.pop();
    for(int i = 1; i <= n; i++) scanf("%d%d", &a[i].l, &a[i].r);
    for(int i = 1; i <= n; i++) a[i].pos = i;
    sort(a + 1, a + n + 1, cmp);
    for(int i = 1; i <= n; i++)  {
        int l = a[i].l, r = a[i].r, pos = a[i].pos;
        if (q.empty() || q.top().r >= l){
            node res;
            res.pos = ++cnt;
            res.l = l;
            res.r = r;
            ans[pos] = cnt;
            q.push(res);
        }
        else {
            node res = q.top();
            q.pop();
            res.l = l;
            res.r = r;
            ans[pos] = res.pos;
            q.push(res);
        }
    }
    printf("%d\n", cnt);
    for(int i = 1; i <= n; i++)  printf("%d\n", ans[i]);
  }
    return 0;
}