Sicily 1033. City Road

最新推荐文章于 2015-03-31 00:24:02 发布

原创最新推荐文章于 2015-03-31 00:24:02 发布 · 417 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#sicily

sicily 专栏收录该内容

413 篇文章

订阅专栏

本文介绍了一个关于城市道路中寻找最短路径数量的问题。在一座由M*N个小正方形房屋组成的城市中，考虑到新建筑可能阻碍某些街道，文章提供了一种算法来重新计算从一个桥到另一个桥的不同最短路径的数量。

1033. City Road

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Long long ago, city Old was all around the water and was divided into M*N small square houses. The city Old had only two bridges, in the southwest corner and northeast corner. It’s obvious that the citizens can have C(m+n, n) different ways with shortest length, to go from one bridge to the other.

As the city developed, bigger buildings came out and blocked some streets. These new buildings were always large rectangles combining many small houses. Since the new buildings may affect the number of shortest ways a lot, it’s you job now to recalculate the number.

The city has M rows and N columns of houses, M+1 horizontal streets, and N+1 vertical streets. Suppose the house in the southwest corner is (1, 1), then the house in the northeast corner can be marked as (m, n). And then each new building can be described by its southwest corner(house) and its size(number of houses) in horizontal and vertical directions.

The picture below shows that the city Old has 5*6 houses and one 3*2 building with its southwest corner located at (2,4). The number of different shortest ways in the picture is 192.

Input

This problem has multiple test cases. In each test case, the first line is two integer m, n(m*n<=1,000,000)which indicate the size of city Old, in vertical and horizontal directions respectively. And the second line has one integer B(B<=1000) what is the number of the new buildings. After that there are B lines. Each line has four integers x, y, a, b, describing the southwest corner (x, y) , the vertical size a, and the horizontal size b of the building. The buildings may be overlapped with each other. The input is terminated by a case with m*n=0, which should not be processed.

Output

For each test case output one integer representing the number of shortest ways in a single line. This number is guaranteed to be less than 2^63.

Sample Input

5 612 4 3 20 0

Sample Output

Problem Source

ZSUACM Team Member

// Problem#: 1033
// Submission#: 3402870
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include <stdio.h>
#include <iostream>
#include <vector>
#include <string>
#include <stack>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <functional>
#include <map>
#include <string.h>
#include <math.h>
using namespace std;

const int MAX = 1000010;

int N, M, B;

bool G[MAX];
long long dp[2][MAX];

int main() {

    //std::ios::sync_with_stdio(false);

    while (1) {

        scanf("%d%d", &M, &N);
        if (!M && !N) break;

        int size = N * M;

        for (int i = 1; i <= size; i++) G[i] = false;

        scanf("%d", &B);
        for (int i = 0; i < B; i++) {
            int x, y, a, b;
            scanf("%d%d%d%d", &x, &y, &a, &b);
            a = x + a - 1;
            b = y + b - 1;
            for (int j = x; j < a; j++) {
                for (int k = y; k < b; k++) {
                    G[(j - 1) * N + k] = true;
                }
            }
        }

        int lastLine = 0, nowLine = 1;

        for (int i = 1; i <= N; i++) {
            dp[0][i] = 1;
            dp[1][i] = 0;
        }

        for (int i = 1; i <= M; i++) {
            dp[nowLine][0] = 1;
            for (int j = 1; j <= N; j++) {
                if (G[(i - 1) * N + j])
                    dp[nowLine][j] = 0;
                else 
                    dp[nowLine][j] = dp[lastLine][j] + dp[nowLine][j - 1];
            }
            nowLine = 1 - nowLine;
            lastLine = 1 - lastLine;
        }

        printf("%lld\n", dp[lastLine][N]);

    }

    return 0;
}