解题思路:
DLX 的模板题,重复覆盖。
#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> #include <assert.h> #define FOR(i,x,y) for(int i=x;i<=y;i++) using namespace std; const int INF = 0x3f3f3f3f; const int maxn = 7500; int boys , boy[100]; struct DLX { #define FF(i,A,s) for(int i = A[s];i != s;i = A[i]) int L[maxn],R[maxn],U[maxn],D[maxn]; int size,col[maxn],row[maxn],s[maxn],H[100]; bool vis[100]; int ans[maxn],cnt; void init(int m) { for(int i=0;i<=m;i++) { L[i] = i - 1;R[i] = i + 1;U[i] = D[i] = i;s[i] = 0; } memset(H,-1,sizeof(H)); L[0] = m;R[m] = 0;size = m + 1; } void link(int r,int c) { U[size] = c;D[size] = D[c];U[D[c]] = size;D[c] = size; if(H[r]<0)H[r] = L[size] = R[size] = size; else { L[size] = H[r];R[size] = R[H[r]]; L[R[H[r]]] = size;R[H[r]] = size; } s[c]++;col[size] = c;row[size] = r;size++; } void del(int c) { L[R[c]] = L[c] ; R[L[c]] = R[c]; FF(i,D,c)FF(j,R,i)U[D[j]] = U[j],D[U[j]] = D[j],--s[col[j]]; } void add(int c) { R[L[c]] = L[R[c]] = c; FF(i,U,c)FF(j,L,i)++s[col[U[D[j]] = D[U[j]] = j]]; } bool dfs(int k) { if(!R[0]) { cnt = k;return 1; } int c = R[0];FF(i,R,0)if(s[c] > s[i])c = i; del(c); FF(i, D, c) { FF(j, R, i) del(col[j]); ans[k] = row[i];if(dfs(k + 1))return true; FF(j,L,i) add(col[j]); } add(c); return 0; } void remove(int c) { FF(i, D, c)L[R[i]] = L[i],R[L[i]] = R[i]; } void resume(int c) { FF(i, U, c)L[R[i]] = R[L[i]] = i; } int A() { int res = 0; memset(vis,0,sizeof(vis)); FF(i, R, 0)if(!vis[i]) { res++;vis[i] = 1; FF(j, D, i)FF(k, R, j)vis[col[k]] = 1; } return res; } void Dance(int now, int &lim) { if(now + A() > boys) return ; int tt = 0; for(int i=0;i<now;i++)tt += boy[ans[i]]; if(now - tt >= lim) return; if(!R[0]) { lim = now - tt; return; } int temp = INF , c; FF(i,R,0)if(temp >= s[i])temp = s[i] , c = i; FF(i, D, c) { ans[now] = row[i]; remove(i);FF(j, R, i)remove(j); Dance(now + 1, lim); FF(j, L, i)resume(j);resume(i); } } }dlx; int N , D; int d[100][100]; int main() { int T , kcase = 1; scanf("%d", &T); while(T--) { scanf("%d%d", &N, &D);dlx.init(N); boys = 0; FOR(i, 1, N) scanf("%d", &boy[i]) , boys += boy[i]; FOR(i, 1, N) FOR(j, 1, N) d[i][j] = (i == j ? 0 : INF); FOR(i, 1, N-1) { int u , v , w; scanf("%d%d%d",&u,&v,&w); d[u][v] = d[v][u] = w; } FOR(k, 1, N) FOR(i, 1, N) FOR(j, 1, N) if(d[i][j] > d[i][k] + d[k][j]) d[i][j] = d[i][k] + d[k][j]; FOR(i, 1, N) FOR(j, 1, N) if(d[i][j] <= D) dlx.link(i, j); int ans = INF; dlx.Dance(0, ans); if(ans > boys) ans = -1; printf("Case #%d: %d\n", kcase++, ans); } return 0; }
时间: 2024-10-12 20:39:31