UVa 10118 Free Candies （记忆化搜索+状态压缩）

来源：互联网发布：如何用c语言编写程序编辑：程序博客网时间：2024/06/07 03:04

题目链接：https://cn.vjudge.net/problem/UVA-10118

思路：设dp[pa][pb][pc][pd]四堆糖分别取到这四堆的第pa、pb、pc、pd颗时最多的pair数，用一个二进制串记录篮中糖的状态（即哪些糖有那些没有）。状态转移一共四个，即分别从这四堆中拿出一颗糖放到篮子里，转移到放之后的状态。详见代码

#include<cstdio>#include<vector>#include<algorithm>#include<map>#include<cmath>#include<cstring>#include<sstream>#include<iostream>using namespace std;const int maxn = 42;int a[maxn], b[maxn], c[maxn], d[maxn]; //four pilesint DP[maxn][maxn][maxn][maxn];int n;inline bool Push(int pre, int m, int &st) // if she can take a pair of candies;{    bool ok = false;    if(pre & (1<<m)) pre ^= (1<<m), ok = true;    else pre |= (1<<m);    st = pre;    return ok;}int dp(int pa, int pb, int pc, int pd, int st, int num) // st : sugars that are in the basket{    int &ans = DP[pa][pb][pc][pd];    if(ans >= 0) return ans;    ans = 0;    if(num < 5) {       if(pa < n) {          int st0;          if(Push(st, a[pa], st0)) ans = max(ans, 1+dp(pa+1, pb, pc, pd, st0, num-1));          else ans = max(ans, dp(pa+1, pb, pc, pd, st0, num+1));       }       if(pb < n) {          int st0;          if(Push(st, b[pb], st0)) ans = max(ans, 1+dp(pa, pb+1, pc, pd, st0, num-1));          else ans = max(ans, dp(pa, pb+1, pc, pd, st0, num+1));       }       if(pc < n) {          int st0;          if(Push(st, c[pc], st0)) ans = max(ans, 1+dp(pa, pb, pc+1, pd, st0, num-1));          else ans = max(ans, dp(pa, pb, pc+1, pd, st0, num+1));       }       if(pd < n) {          int st0;          if(Push(st, d[pd], st0)) ans = max(ans, 1+dp(pa, pb, pc, pd+1, st0, num-1));          else ans = max(ans, dp(pa, pb, pc, pd+1, st0, num+1));       }    }    return ans;}int main(){   while(scanf("%d", &n) && n) {      memset(DP, -1, sizeof DP);      for(int i = 0; i < n; i++)         scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);      printf("%d\n", dp(0, 0, 0, 0, 0, 0));   }}

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