ZOJ1894 POJ2465 Adventures in Moving - Part IV

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典型的DP题，dp[i][j]表示的是到了第i个站的时候，总共油缸里还剩下j油的最小费用。记得把起始和终点都要列进去。

输入真恶心

/******************************************************************************* # Author : Neo Fung # Email : neosfung@gmail.com # Last modified: 2012-04-16 21:41 # Filename: ZOJ1894 POJ2465 Adventures in Moving - Part IV.cpp # Description :  ******************************************************************************/#ifdef _MSC_VER#define DEBUG#define _CRT_SECURE_NO_DEPRECATE#endif#include <fstream>#include <stdio.h>#include <iostream>#include <string.h>#include <string>#include <limits.h>#include <algorithm>#include <math.h>#include <numeric>#include <functional>#include <ctype.h>#define MAX 110#define inf 20000000using namespace std;int dp[MAX][MAX*2],dis[MAX],price[MAX],total;int main(void){#ifdef DEBUG  freopen("../stdin.txt","r",stdin);freopen("../stdout.txt","w",stdout); #endif  char str[100];while(~scanf("%d",&total)){int n=1;getchar();while(gets(str) && strlen(str)){sscanf(str,"%d%d",&dis[n],&price[n]);++n;}dis[n]=total;price[n++]=inf;for(int i=0;i<MAX;++i)for(int j=0;j<MAX*2;++j)dp[i][j]=inf;dp[0][100]=0;for(int i=1;i<n;++i)for(int j=0;j<=200;++j){for(int k=dis[i]-dis[i-1];k<=200;++k)if(dp[i-1][k]!=inf){if(k-(dis[i]-dis[i-1])>j)break;elsedp[i][j]=min(dp[i][j],dp[i-1][k]+(j-(k-dis[i]+dis[i-1]))*price[i]);}}int ans=dp[n-1][100];if(ans==inf)printf("Impossible\n");elseprintf("%d\n",ans);}return 0;}