[UVA][dp] 11951 - Area＠Morris' Blog｜PChome Online 人新台

Morris' Blog 我在摸索我存在的意、生命的意、涵及值。有真正神手般的害，套用模式是大部分的情。那害的我存在的意何，托出神的存在？有候不能怪人不我，我已始省思，什我有人的原因，而是全要靠人。有候我我不同，那是因我做不到很多人能到的事情。我不到、玩不起、得不明、理解得不多，我缺少的文能力，就是一切一切能的判定。此，我不得能做些人有助的事情。我知道我自己是不服的性格，但在我不得不。我有能力，因此我必。我到底世界作什？充不定性，而我是法出拔萃，想一般人似乎也回不了了。越大的挑需要越力的支持，而我的支持是什？最近人，但都是失的局，人能接受，不需要言，不需要通，一切所想都表不出。
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2013-03-24 22:37:20| 人1,095| 回0 | 上一篇 | 下一篇

[UVA][dp] 11951 - Area

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A Area

Time Limit: 1 sec
Memory Limit: 32 MB

Steve has a dream to move from dormitory where we live now to a private house. He has already earned some money by solving various problems and now wants to make a first move towards his dream. Steve is going to buy a plot for his future house. However, due to high tax rate, he is going to buy only one plot. Certainly Steve wants to maximize its area for the amount of money he owns (and if there are several plots of same area, Steve certainly chooses a cheaper one). Can you help Steve?

The whole district where Steve wants to reside is divided into N * M equal strips. Each strip is sold separately, for its own price P_i;j. According to the country law, plot is defined as a rectangular collection of strips with sides parallel to the district.

INPUT

There is a number of tests T (T ≤ 110) on the first line. After T tests follows. Each test case is defined by three numbers N M K (N; M ≤ 100; K ≤ 10⁹). Next N lines with M numbers each stands for prices of the strips P_i;j (P_i;j ≤ 10⁶).

OUTPUT

For each test print a line formatted as "Case #T: S P", where T stands for test case number (starting from 1), S for maximal plot area that Steve can afford and P for plot’s total cost.

SAMPLE INPUT

1 5 6 15 10 1 10 20 10 10 30 1 1 5 1 1 50 1 1 20 1 1 10 5 5 10 5 1 40 10 90 1 10 10

SAMPLE OUTPUT

Case #1: 6 10

Problem by: Aleksej Viktorchik; Leonid Sislo
Huge Easy Contest #2

O(n^3)

#include <stdio.h>
inline int readchar() {
    const int N = 1048576;
    static char buf[N];
    static char *p = buf, *end = buf;
    if(p == end) {
        if((end = buf + fread(buf, 1, N, stdin)) == buf) return EOF;
        p = buf;
    }
    return *p++;
}
inline int ReadInt(int *x) {
    static char c, neg;
    while((c = readchar()) < '-')    {if(c == EOF) return 0;}
    neg = (c == '-') ? -1 : 1;
    *x = (neg == 1) ? c-'0' : 0;
    while((c = readchar()) >= '0')
        *x = (*x << 3) + (*x << 1) + c-'0';
    *x *= neg;
    return 1;
}
int main() {
    int t, N, M, K;
    int g[105][105];
    int cases = 0;
    int i, j, k, l;
    //scanf("%d", &t);
    ReadInt(&t);
    while(t--) {
        //scanf("%d %d %d", &N, &M, &K);
        ReadInt(&N);
        ReadInt(&M);
        ReadInt(&K);
        for(i = 0; i < N; i++)
            for(j = 0; j < M; j++)
                //scanf("%d", &g[i][j]);
                ReadInt(&g[i][j]);
        int mx = 0, mxv = 0;
  &nbs;     long long tmp;
        int cnt = 0;
        for(i = 0; i < N; i++) {
            int sum[105] = {};
            for(j = i; j < N; j++) {
                for(k = 0, l = 0; k < M; k++) {
                    sum[k] += g[j][k];
                    if(k == 0) tmp = 0;
                    tmp += sum[k];
                    while(tmp > K)
                        tmp -= sum[l++];
                    if((j-i+1)*(k-l+1) > mx)
                        mx = (j-i+1)*(k-l+1), mxv = K;
                    if((j-i+1)*(k-l+1) == mx && tmp < mxv)
                        mxv = tmp;
                }
            }
        }
        printf("Case #%d: %d %d\n", ++cases, mx, mxv);
    }
    return 0;
}