[BFS]HDU1044 Collect More Jewels-优快云博客

本文链接：https://blog.youkuaiyun.com/bit_Line/article/details/45245813

本文详细介绍了使用两遍BFS算法解决HDU1044问题的过程，包括距离记录、搜索策略及估值函数设置，旨在帮助读者理解并应用此算法解决类似问题。

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HDU1044 Collect More Jewels
代码

HDU1044 Collect More Jewels

两遍BFS，首先需要记录一下每个点到DEST的距离，然后就可以搜了。当然也可以直接A*搞一下，估值函数设为f(x)={dis<

代码:

#include <cstdio>
#include <cstring>
#include <cctype>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;

const int MAX = 51;

int W, H, L, M;
int fetch[MAX];
char str[MAX][MAX];
bool vis[MAX][MAX][1 << 11];
int dis[MAX][MAX];
struct node {
    int x, y;
    int value;
    int vis;
    int money;
} S, T;
int d[4][2] = {-1,0, 0,1, 1,0, 0,-1};

inline bool valid(int x, int y) {
    return x >= 0 && y >= 0 && x < H && y < W && str[x][y] != '*';
}

void BFS1(const node& S) {
    queue<node> Q;
    node p, q;
    Q.push(S);
    memset(dis, -1, sizeof(dis));
    dis[S.x][S.y] = 0;
    while (!Q.empty()) {
        p = Q.front();
        Q.pop();

        q.value = p.value + 1;
        for (int i = 0; i < 4; ++i) {
            q.x = p.x + d[i][0];
            q.y = p.y + d[i][1];
            if (valid(q.x, q.y) && dis[q.x][q.y] == -1) {
                dis[q.x][q.y] = q.value;
                Q.push(q);
            }
        }
    }
}

int BFS2(const node& S, const node& T) {
    memset(vis, false, sizeof(vis));
    vis[S.x][S.y][S.vis] = true;
    queue<node> Q;
    node p, q;
    Q.push(S);
    int ans = -1;
    while (!Q.empty()) {
        p = Q.front();
        //printf("fetch(%d,%d)\n", p.x, p.y);
        Q.pop();

        q.value = p.value + 1;
        for (int i = 0; i < 4; ++i) {
            q.x = p.x + d[i][0];
            q.y = p.y + d[i][1];
            if (valid(q.x, q.y)) {
                //printf("1:(%d,%d)\n", q.x, q.y);
                if (q.value + dis[q.x][q.y] <= L) {
                    //printf("2:(%d,%d)\n", q.x, q.y);
                    if (isalpha(str[q.x][q.y])) {
                        int tp = 1 << (str[q.x][q.y] - 'A');
                        q.vis = p.vis | tp;
                        q.money = p.money + ((p.vis & tp) ? 0 : fetch[str[q.x][q.y] - 'A']);
                    } else {
                        q.vis = p.vis;
                        q.money = p.money;
                    }
                    if (!vis[q.x][q.y][q.vis]) {
                        //printf("3:(%d,%d)\n", q.x, q.y);
                        vis[q.x][q.y][q.vis] = true;
                        //printf("vis(%d, %d, %d, %d)\n", q.x, q.y, q.vis, q.money);
                        Q.push(q);
                        if (q.x == T.x && q.y == T.y) {
                            ans = max(ans, q.money);
                        }
                    }
                }
            }
        }
    }
    return ans;
}

int main() {
    int cas;
    scanf(" %d", &cas);
    for (int t = 1; t <= cas; ++t) {
        scanf(" %d %d %d %d", &W, &H, &L, &M);
        for (int i = 0; i < M; ++i) {
            scanf(" %d", fetch + i);
        }
        for (int i = 0; i < H; ++i) {
            scanf(" %s", str[i]);
            for (int j = 0; j < W; ++j) {
                if (str[i][j] == '@') {
                    S.x = i;
                    S.y = j;
                    S.value = S.vis = S.money = 0;
                    str[i][j] = '.';
                } else if (str[i][j] == '<') {
                    T.x = i;
                    T.y = j;
                    T.value = 0;
                    str[i][j] = '.';
                }
            }
        }
        BFS1(T);
        int ans = BFS2(S, T);
        if (t > 1) puts("");
        printf("Case %d:\n", t);
        if (ans < 0) {
            puts("Impossible");
        } else {
            printf("The best score is %d.\n", ans);
        }
    }
    return 0;
}