HOJ Piggy-Bank 为什么？？？？

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Piggy-Bank
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 Source : ACM ICPC Central European Regional 1999 Time limit : 5 sec Memory limit : 32 M
Submitted : 1216, Accepted : 517
Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action comes from Irreversibly Bound Money (IBM). The idea behind is simple. Whenever some ACM member has any small money, he takes all the coins and throws them into a piggy-bank. You know that this process is irreversible, the coins cannot be removed without breaking the pig. After a sufficiently long time, there should be enough cash in the piggy-bank to pay everything that needs to be paid.
But there is a big problem with piggy-banks. It is not possible to determine how much money is inside. So we might break the pig into pieces only to find out that there is not enough money. Clearly, we want to avoid this unpleasant situation. The only possibility is to weigh the piggy-bank and try to guess how many coins are inside. Assume that we are able to determine the weight of the pig exactly and that we know the weights of all coins of a given currency. Then there is some minimum amount of money in the piggy-bank that we can guarantee. Your task is to find out this worst case and determine the minimum amount of cash inside the piggy-bank. We need your help. No more prematurely broken pigs!
Input Specification
The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing two integers E and F. They indicate the weight of an empty pig and of the pig filled with coins. Both weights are given in grams. No pig will weigh more than 10 kg, that means 1 <= E <= F <= 10000. On the second line of each test case, there is an integer number N (1 <= N <= 500) that gives the number of various coins used in the given currency. Following this are exactly N lines, each specifying one coin type. These lines contain two integers each, Pand W (1 <= P <= 50000, 1 <= W <=10000). P is the value of the coin in monetary units, W is it's weight in grams.
Output Specification
Print exactly one line of output for each test case. The line must contain the sentence "The minimum amount of money in the piggy-bank is X. where X is the minimum amount of money that can be achieved using coins with the given total weight. If the weight cannot be reached exactly, print a line "This is impossible..
Sample Input:
310 11021 130 5010 11021 150 301 6210 320 4
Sample Output:
The minimum amount of money in the piggy-bank is 60.The minimum amount of money in the piggy-bank is 100.This is impossible.
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Copyright © 1998-2011 HIT ACM/ICPC Group
All Rights Reserved.

/* This Code is Submitted by 陈玉标 for Problem 1031 at 2011-07-31 21:38:18 */
#include <iostream>
using namespace std;
const int MAX = 10001;
const int largest = 1000000;
int dp[MAX];
int w[501];
int p[501];
int N;


/*
 * 可以小试牛刀。终于没参考被人的代码写了个水题。
 */
void DP(int m) {
    for (int i = 1; i <= m; i++) {
        dp[i] = largest;
    }
    dp[0] = 0;
    for (int i = 1; i <= N; i++) {
        for (int j = w[i]; j <= m; j++) {
            if (dp[j - w[i]] + p[i] < dp[j]) {
                dp[j] = dp[j - w[i]] + p[i];
            }
        }
    }
}


int main(int argc, char** argv) {
    int T, E, F, W;
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d%d", &E, &F, &N);
        for (int i = 1; i <= N; i++) {
            scanf("%d%d", &p[i], &w[i]);
        }
        W = F - E;
        DP(W);
        if (dp[W] == largest) {
            printf("This is impossible.\n");
        } else {
            printf("The minimum amount of money in the piggy-bank is %d.\n", dp[W]);
        }
    }
    return 0;
}


过不了的代码：
/*
 * File:   main.cpp
 * Author: Administrator
 *
 * Created on 2012年3月21日, 下午8:08
 */




#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int INF = 1000000;
const int MAXW = 10010;
const int MAXN = 510;
int P[MAXN];
int W[MAXN];
int dp[MAXN][MAXW];
int E, F, N, V;




/*
 * 状态转移方程为：dp[i][v]=min{dp[i-1][v],dp[i-1][v-c[i]]+w[i]};
 * 对于不存在的状态，为了不影响min，都应该是INF,这个是可以状态压缩的
 * 为什么一维转化为二维就错了，谁能指点
 */




void DP() {
    
    for (int i = 0; i <= N; i++) {
        for (int j = 0; j <= V; j++) {
            dp[i][j] = INF;
        }
    }
    for (int i = 0; i <= N; i++) {
        dp[i][0] = 0;
    }




    for (int i = 1; i <= N; i++) {
        for (int j = W[i]; j <= V; j++) {
            if (dp[i - 1][j] < (dp[i - 1][j - W[i]] + P[i])) {
                dp[i][j] = dp[i - 1][j];
            } else {
                dp[i - 1][j] = dp[i][j] = (dp[i - 1][j - W[i]] + P[i]);
            }
        }
    }




    if (dp[N][V] == INF) {
        printf("This is impossible.\n");
    } else {
        printf("The minimum amount of money in the piggy-bank is %d.\n", dp[N][V]);
    }
}




int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d%d", &E, &F, &N);
        for (int i = 1; i <= N; i++) {
            scanf("%d%d", &P[i], &W[i]);
        }
        V = F - E;
        DP();
    }