Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

The researchers at International Bio-bot Makers (IBM) have invented a new kind of Bio-bot, a robot with behavior mimicking biological organisms. The development of the new robot is at a primitive stage; they now resemble simple four-wheeled rovers. And like most modern robots, Bio-bots are not very mobile. Their weak motors and limited turning capability put considerable limitations on their movement, even in simple, relatively obstacle-free environments.

Currently the Bio-bots operate in a room which can be described as an m x n grid. A Bio-bot occupies a full square on this grid. The exit is in the northeast corner, and the room slopes down towards it, which means the Bio-bots are only capable of moving north or east at any time. Some squares in the room are also occupied by walls, which completely block the robot. Figure 1, which corresponds to the sample input, shows an example of such a room.

Clearly, a Bio-bot located on square A is capable of leaving the room, while one at square B is trapped inside it, no matter what it does. Locations like B are called ``stuck squares." (Walls do not count as stuck squares.) Given the description of a room, your job is to count the total number of stuck squares in the room.

Input 

Input consists of multiple test cases, each describing one room. Each test case begins with a line containing three integers m, n, and w (1m, n10⁶, 0w1000). These indicate that the room contains m rows, n columns, and w horizontal walls.

Each of the next w lines contains four integers x₁, y₁, x₂, y₂, the coordinates of the squares delimiting one wall. All walls are aligned from west to east, so 0x₁x₂ < n and 0y₁ = y₂ < m. Walls do not overlap each other. The southwest corner of the room has coordinates (0,0) and the northeast corner has coordinates (n - 1, m - 1).

The last test case is followed by a line containing three zeros.

Output 

For each test case, display one line of output containing the test case number followed by the number of stuck squares in the given room. Follow the format shown in the sample output.

Sample Input 

8 8 3 
1 6 3 6 
2 4 2 4 
4 2 7 2 
0 0 0

Sample Output 

Case 1: 8

---

題目描述：

只能往右往上走，求圖中有多少點無法抵達右上角。

題目解法：

首先，先將 x, y 軸離散化，用 dp 推回去找到可以抵達的點。

對於不能抵達的點，將離散化的結果拿出來計算該點有多少格子。

#include <stdio.h>
#include <string.h>
#include <map>
#include <algorithm>
using namespace std;
char g[2010][2010], dp[2010][2010];
int lx[4010], rx[4010], ly[4010], ry[4010];
long long x[4010], y[4010];
int main() {
    //freopen("in.txt","r+t",stdin);
    //freopen("out.txt","w+t",stdout);
    int cases = 0;
    int m, n, w;
    int i, j, k, p, q, r;
    while(scanf("%d %d %d", &m, &n, &w) == 3 && n+m) {
        map<int, int> Rx, Ry;
        Rx[0] = 1, Ry[0] = 1;
        Rx[n] = 1, Ry[m] = 1;
        for(i = 0; i < w; i++) {
            scanf("%d %d %d %d", &lx[i], &ly[i], &rx[i], &ry[i]);
            rx[i]++, ry[i]++;
            Rx[lx[i]] = 1, Ry[ly[i]] = 1;
            Rx[rx[i]] = 1, Ry[ry[i]] = 1;
        }
        int sizex = 0, sizey = 0;
        for(map<int, int>::iterator it = Rx.begin();
            it != Rx.end(); it++) {
            it->second = sizex;
            x[sizex++] = it->first;
        }
        for(map<int, int>::iterator it = Ry.begin();
            it != Ry.end(); it++) {
            it->second = sizey;
            y[sizey++] = it->first;
        }
        memset(g, 0, sizeof(g));
        memset(dp, 0, sizeof(dp));
        for(i = 0; i < w; i++) {
            int llx = Rx[lx[i]], rrx = Rx[rx[i]];
            int rry = Ry[ry[i]];
            for(p = llx+1; p <= rrx; p++)
                g[p][rry] = 1;
        }
        dp[sizex-1][sizey-1] = 1;
        long long ret = 0;
        for(i = sizex-1; i > 0; i--) {
            for(j = sizey-1; j > 0; j--) {
                if(g[i][j])
                    continue;
                if(!dp[i][j])
                    ret += (x[i]-x[i-1])*(y[j]-y[j-1]);
                dp[i-1][j] |= dp[i][j];
                dp[i][j-1] |= dp[i][j];
            }
        }
        printf("Case %d: %lld\n", ++cases, ret);
    }
    return 0;
}