POJ 1018 Communication System

WA 的分析

1. solve_dp 返回的是 double 不是 int 浪费了非常久的时间

2. 做动规, 当状态转移方程比较复杂时, 举例子能够帮助初始化和写出状态转移方程

 

总结

1. 这里 dp 的设置是 tight, 一般 tight 都是比较难的题目

 

#include <iostream>
#include <stdio.h>
#include <memory.h>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <deque>
#include <cstring>
#define MIN(x,y) (x)<(y)?(x):(y)
#define MAX(x,y) (x)>(y)?(x):(y)
using namespace std;

vector<int> money[200];
vector<int> bandWidth[200];

int dp[200][200];

// tight, not exactly 
// handle to decide the iterator function
// use examples to help understand
double solve_dp(int n, int maxbw)  {
	memset(dp, 0x3f, sizeof(dp));
	
	// init
	dp[0][0] = 0;
	for(int i = 0; i < money[0].size(); i ++)  {
		for(int j = 1; j <= bandWidth[0][i]; j ++)  {
			dp[0][j] = min(dp[0][j], money[0][i]);
		}
	}

	// main procedure
	for(int i = 1; i < n; i ++)  {
		for(int  j = 0; j <= maxbw; j ++)  {
			bool init = false;
			for(int k = 0; k < money[i].size(); k ++)  {
				if(j > bandWidth[i][k]) continue;
				if(init)  {
					dp[i][j] = dp[i-1][j] + money[i][k];
					init = true;
				}  else  {
					dp[i][j] = min(dp[i][j], dp[i-1][j] + money[i][k]);
				}
					
			}
		}
	}

	double res = 0.0;
	for(int i = 1; i <= maxbw; i ++)  {
		if(dp[n-1][i] == 0x3f3f3f3f) continue;
		res = MAX(res, i*1.0/dp[n-1][i]);
	}

	return res;
}

int main() {
	freopen("C:\\Users\\vincent\\Dropbox\\workplacce\\joj\\test.txt", "r", stdin);
	
	int T, N;
	scanf("%d", &T);
	while(T--)  {
		scanf("%d", &N);

		int maxbw = 0;
		for(int i = 0; i < N; i ++)  {
			int number;
			scanf("%d", &number);
			money[i].clear();
			bandWidth[i].clear();
			for(int j = 0; j < number; j ++)  {
				int bw, pc;
				scanf("%d%d", &bw, &pc);
				bandWidth[i].push_back(bw);
				money[i].push_back(pc);
				maxbw = max(maxbw, bw);
			}
		}

		double res = solve_dp(N, maxbw);
		printf("%0.3f\n", res);
	}

	return 0;
}