POJ 2923 Relocation(状态压缩DP)

题目描述

 

总结

1. 这种状态转移的题目不太好 debug, 写起来总觉得不对劲

2. 动规的初始化总能找到一些简练的初始化方法, 比如这道题, 可以选择 j>=0, dp[0] = 1, 或者对 vector 中的所有元素赋值 dp[s] = 1, 然后 j > 0

 

代码 TLE 了

/*
 * source.cpp
 *
 *  Created on: Apr 6, 2014
 *      Author: sangs
 */


#include <stdio.h>
#include <iostream>
#include <string>
#include <vector>
#include <memory.h>
#include <algorithm>
#include <math.h>
using namespace std;

int arr[100];

vector<int> state;
int dp[100100];

bool cal(int n, int si, int load1, int load2) {
	int sum = 0;
	bool dp1[200];
	memset(dp1, 0, sizeof(dp1));
	dp1[0] = true;


	for(int i = 0; i < n; i ++) {
		if(((1<<i) & si )== 0)
			continue;
		sum += arr[i];
		if(arr[i] > load1) continue;

		for(int j = load1; j >= arr[i]; j -- ) {
			dp1[j] |= dp1[j-arr[i]];
		}
	}

	int sum1 = 0;
	for(int i = load1; i >= 0; i --) {
		if(dp1[i]) {
			sum1 = i;
			break;
		}
	}
	if(sum-sum1 <= load2)
		return true;
	return false;
}

void func1(int n, int load1, int load2) {
	int endState = (1<<n);
	for(int i = 1; i < endState; i++) {
		if(cal(n, i, load1, load2)) {
			state.push_back(i);
		}
	}
}


int solve_dp(int n) {

	memset(dp, 0x3f, sizeof(dp));

	dp[0] = 0;
	for(size_t i = 0; i < state.size(); i ++) {
		for(int j = (1<<n)-1; j >= 0; j --) {
			dp[j|state[i]] = min(dp[j|state[i]], dp[j]+1);
		}
	}

	return dp[(1<<n)-1];
}

int main() {
	freopen("input.txt", "r", stdin);
	int cases;
	scanf("%d", &cases);
	int seq = 0;
	while(cases --) {
		seq ++;

		int n, load1, load2;
		scanf("%d%d%d", &n, &load1, &load2);

		for(int i = 0; i < n; i ++)
			scanf("%d", arr+i);
		func1(n, load1, load2);
		int res = solve_dp(n);

		printf("Scenario #%d:\n%d\n\n", seq, res);
	}
	return 0;
}