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

Problem B: Ocean Currents

For a boat on a large body of water, strong currents can be dangerous, but with careful planning, they can be harnessed to help the boat reach its destination. Your job is to help in that planning.

At each location, the current flows in some direction. The captain can choose to either go with the flow of the current, using no energy, or to move one square in any other direction, at the cost of one energy unit. The boat always moves in one of the following eight directions: north, south, east, west, northeast, northwest, southeast, southwest. The boat cannot leave the boundary of the lake. You are to help him devise a strategy to reach the destination with the minimum energy consumption.

Input Specification

The lake is represented as a rectangular grid. The first line of input contains two integers r and c, the number of rows and columns in the grid. The grid has no more than 1000 rows and no more than 1000 columns. Each of the following r lines contains exactly c characters, each a digit from 0 to 7 inclusive. The character 0 means the current flows north (i.e. up in the grid, in the direction of decreasing row number), 1 means it flows northeast, 2 means east (i.e. in the direction of increasing column number), 3 means southeast, and so on in a clockwise manner:

7 0 1
 \|/
6-*-2
 /|\
5 4 3

The line after the grid contains a single integer n, the number of trips to be made, which is at most 50. Each of the following n lines describes a trip using four integers rs, cs, rd, cd, giving the row and column of the starting point and the destination of the trip. Rows and columns are numbered starting with 1.

Sample Input

5 5
04125
03355
64734
72377
02062
3
4 2 4 2
4 5 1 4
5 3 3 4

Output Specification

For each trip, output a line containing a single integer, the minimum number of energy units needed to get from the starting point to the destination.

Output for Sample Input

0
2
1

---

Ondřej Lhoták, Richard Peng

題目描述；
盤面上有海流，而被海流沖向的方位取決於當前那格的數字。
如果不順著海流走都會消耗 1 能量，問最少的能量從起點到終點。

基本上就是 priority_queue, 加了一個估價函數, 但只是希望在相同的時候以
最接近的那個抓出來拓展。
如果用 spfa 的話可能會更新所有盤面，不知成效如何。


#include <stdio.h>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
char g[1005][1005];
short used[1005][1005];
short dist[1005][1005];
int n, m;
struct E {
    short x, y, d, h;
    E(short a, short b, short c, short dd):
        x(a), y(b), d(c), h(dd){}
    bool operator<(const E &a) const {
        if(d != a.d)
            return d > a.d;
        return h > a.h;
    }
};
priority_queue<E, vector<E> > pQ;
void sol(int sx, int sy, int ex, int ey) {
    static int times = 0;
    times++;
    while(!pQ.empty())  pQ.pop();
    E tn(0,0,0,0);
    pQ.push(E(sx, sy, 0, ex-sx+ey-sy));
    int dx[] = {-1,-1,0,1,1,1,0,-1};
    int dy[] = {0,1,1,1,0,-1,-1,-1};
    int i, cost, tx, ty, dir;
    used[sx][sy] = times;
    dist[sx][sy] = 0;
    while(!pQ.empty()) {
        tn = pQ.top(), pQ.pop();
        if(tn.x == ex && tn.y == ey) {
            printf("%d\n", tn.d);
            break;
        }
        dir = g[tn.x][tn.y] -'0';
        short &o = dist[tn.x][tn.y];
        for(i = 0; i < 8; i++) {
            tx = tn.x + dx[i], ty = tn.y + dy[i];
            if(tx < 0 || tx >= n || ty < 0 || ty >= m)
                continue;
            cost = 0;
            if(dir != i)
                cost = 1;
            if(dist[tx][ty] > o+cost || used[tx][ty] != times) {
                used[tx][ty] = times;
                dist[tx][ty] = o+cost;
                pQ.push(E(tx, ty, dist[tx][ty], ex-tx+ey-ty));
            }
        }
    }
}
int main() {
    int i, q, sx, sy, ex, ey;
    while(scanf("%d %d", &n, &m) == 2) {
        while(getchar() != '\n');
        for(i = 0; i < n; i++)
            gets(g[i]);
        scanf("%d", &q);
        while(q--) {
            scanf("%d %d %d %d", &sx, &sy, &ex, &ey);
            sx--, sy--, ex--, ey--;
            sol(sx, sy, ex, ey);
        }
    }
    return 0;
}