标签:
状态压缩的DP,dp[i][st]表示状态为st考虑后面i个人所有人最小花费,
因为每个科目有三种状态,可以用一个三进制数表示,
状态不是很多,所以可以把预先把每个数的三进制预处理出来,
决策为选和不选。
#include<bits/stdc++.h> using namespace std; const int maxs = 8; const int MAXSTA = 6561+5; const int maxn = 101; int base[maxs+1],trp[MAXSTA][maxs+1]; int dp[maxn][MAXSTA]; int tch[maxn],cost[maxn],sa; int s,m,n; const int INF = 0x3fffffff; void preDeal() { base[0] = 1; for(int i = 1; i <= maxs; i++){ base[i] = 3*base[i-1]; } for(int i = 1; i < MAXSTA; i++){ int x = i, cnt = 0; while(x){ trp[i][cnt++] = x%3; x /= 3; } } } int dfs(int i,int st) { if(i == n) return st == sa? 0:INF; int &ans = dp[i][st]; if(~ans) return ans; ans = dfs(i+1,st); //int nst = st; for(int j = 0; j < s; j++){ if(trp[tch[i]][j] && trp[st][j] != 2){ st += base[j]; } } ans = min(ans,dfs(i+1,st)+cost[i]); return ans; } int work() { sa = base[s]-1; int st = 0,sum = 0; for(int i = 0; i < m; i++){ int c; scanf("%d",&c); sum += c; while(getchar()!=‘\n‘){ int sub = getchar()-‘1‘; int t = trp[st][sub] ; if(t != 2){ st += base[sub]; } } } for(int i = 0; i < n; i++){ scanf("%d",cost+i); tch[i] = 0; while(getchar()!=‘\n‘){ tch[i] += base[getchar()-‘1‘]; } } memset(dp,-1,sizeof(dp)); return sum + dfs(0,st); } int main() { //freopen("in.txt","r",stdin); preDeal(); while(scanf("%d%d%d",&s,&m,&n),s){ printf("%d\n",work()); } return 0; }
UVA 10817 - Headmaster's Headache
标签:
原文地址:http://www.cnblogs.com/jerryRey/p/4746428.html