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HDU 4756 Install Air Conditioning (MST+树形DP)

题意:n-1个宿舍,1个供电站,n个位置每两个位置都有边相连,其中有一条边不能连,求n个位置连通的最小花费的最大值。

析:因为要连通,还要权值最小,所以就是MST了,然后就是改变一条边,然后去找出改变哪条能使得总花费最大,dp[i][j] 表示那条边左边的 i 和右边的 j,

最短距离,然后枚举MST里面的每条边,就能知道哪是最大了,注意 供电站和宿舍之间的边不能考虑的。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>//#include <tr1/unordered_map>#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;//using namespace std :: tr1;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e3 + 5;const LL mod = 10000000000007;const int N = 1e6 + 5;const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){    return r >= 0 && r < n && c >= 0 && c < m;}struct Point{    double x, y;};struct Edge{    int to, next;};Edge edge[maxn<<1];Point a[maxn];double dist[maxn][maxn], lowc[maxn], dp[maxn][maxn];bool vis[maxn], is_tree[maxn][maxn];int pre[maxn], head[maxn];int cnt;double sum, ans;double Distan(const Point& lhs, const Point& rhs){    return sqrt((lhs.x - rhs.x) * (lhs.x - rhs.x) + (lhs.y - rhs.y) * (lhs.y - rhs.y));}void add(int u, int v){    edge[cnt].to = v;    edge[cnt].next = head[u];    head[u] = cnt++;}void Prim(){    sum = 0.0;    memset(vis, false, sizeof vis);    memset(pre, 0, sizeof pre);    for(int i = 1; i < n; ++i)  lowc[i] = dist[0][i];    vis[0] = true;    for(int i = 1; i < n; ++i){        double minc = inf;        int p = -1;        for(int j = 0; j < n; ++j)            if(!vis[j] && minc > lowc[j]) minc = lowc[j], p = j;        sum += minc;        vis[p] = true;        add(p, pre[p]);        add(pre[p], p);        for(int j = 0; j < n; ++j)            if(!vis[j] && lowc[j] > dist[p][j])                lowc[j] = dist[p][j], pre[j] = p;    }}double dfs(int u, int fa, int root){    double ans = fa == root ? inf : dist[root][u];    for(int i = head[u]; ~i; i = edge[i].next){        int v = edge[i].to;        if(v == fa)  continue;        double tmp = dfs(v, u, root);        ans = min(ans, tmp);        dp[u][v] = dp[v][u] = min(dp[u][v], tmp);    }    return ans;}void dfs1(int u, int fa){    for(int i = head[u]; ~i; i = edge[i].next){        int v = edge[i].to;        if(v == fa)  continue;        if(fa)  ans = max(ans, sum-dist[u][v]+dp[u][v]);        dfs1(v, u);    }}int main(){    int T;  cin >> T;    while(T--){        scanf("%d %d", &n, &m);        for(int i = 0; i < n; ++i)  scanf("%lf %lf", &a[i].x, &a[i].y);        for(int i = 0; i < n; ++i)            for(int j = i+1; j < n; ++j)                dist[i][j] = dist[j][i] = Distan(a[i], a[j]);        cnt = 0;        memset(head, -1, sizeof head);        Prim();        for(int i = 0; i < n; ++i)            for(int j = 0; j < n; ++j)  dp[i][j] = inf;        for(int i = 0; i < n; ++i)  dfs(i, -1, i);        ans = sum;        dfs1(0, 0);        ans *= m * 1.0;        printf("%.2f\n", ans);    }    return 0;}

 

HDU 4756 Install Air Conditioning (MST+树形DP)