hdu1044(bfs+dfs/bfs+状态压缩)

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1044
题目大意：
在一个迷宫中，从起点走到终点，还有几个宝物，问在给定的时间内，到达终点后所能获取的最大价值。

1.状态压缩记录状态，用十位的二进制数表示每个宝石选还是不选，共有2^10=1024个状态,开个数组vis[maxn][maxn][1024]判断在每一点是否达到该状态，bfs搜索最短路径，W*H*1024个状态都要搜到，复杂度较高

#include <stdio.h>#include <queue>#include <memory.h>using namespace std;int W, H, L, M;bool vis[55][55][1100];char g[55][55];int money[15];int dx[] = { -1, 1, 0, 0 };int dy[] = { 0, 0, -1, 1 };struct node{    int x, y, state, val, time;};bool check(int x, int y){    if (x<0 || x>H - 1 || y<0 || y>W - 1)return false;    if (g[x][y] == '*')return false;    return true;}int bfs(int sx, int sy){    memset(vis, 0, sizeof(vis));    queue<node>q;    node cur, next;    cur.x = sx;    cur.y = sy;    cur.state = 0;    cur.time = 0;    cur.val = 0;    q.push(cur);    vis[sx][sy][0] = 1;    int ans = -1;    while (!q.empty()){        cur = q.front();        q.pop();        for (int i = 0; i < 4; ++i){            next.x = cur.x + dx[i];            next.y = cur.y + dy[i];            if (check(next.x, next.y)){                if (g[next.x][next.y] >= 'A'&&g[next.x][next.y] <= 'J'){                    int tmp = g[next.x][next.y] - 'A';                    if (!(cur.state&(1<<tmp))){//这一句WA到死，不仅要判断状态是否到达过，而且每个宝石只能拿一次啊，101和100不是同一个状态，但拿了同一个宝石                        next.state = cur.state | (1 << tmp);                        next.val = cur.val + money[tmp];                    }                }                else{                    next.state = cur.state;                    next.val = cur.val;                }                if (!vis[next.x][next.y][next.state]){                    next.time = cur.time + 1;                    vis[next.x][next.y][next.state] = 1;                    if (g[next.x][next.y] == '<'){                        ans = max(ans, next.val);                    }                    if (next.time == L)continue;                    q.push(next);                                }            }                                }    }    return ans;}int main(){    //freopen("in.txt", "r", stdin);    int T, sx, sy;    scanf("%d", &T);    for (int _case = 1; _case <= T; ++_case){        scanf("%d %d %d %d", &W, &H, &L, &M);        for (int i = 0; i < M; ++i)scanf("%d", &money[i]);        for (int i = 0; i < H; ++i)        {            scanf("%s", g[i]);            for (int j = 0; j < W; ++j)            {                if (g[i][j] == '@')                {                    sx = i;                    sy = j;                    g[i][j] == '.';                }            }        }        printf("Case %d:\n", _case);        int ans = bfs(sx, sy);        if (ans >= 0)printf("The best score is %d.\n", ans);        else printf("Impossible\n");        if (_case < T)printf("\n");    }    return 0;}

2.bfs+dfs

bfs找到出口到每个宝石，每个宝石之间和所有宝石到出口的路径，只用搜索每个x,y是否已经到达过，复杂度大大降低

dfs找到在time limit内是否能找到出口，记录能拿到的宝石的最大值

注意dfs的剪枝，如果此时拿到宝石的数量已经达到总量，可以不用继续搜索下去了

#include<stdio.h>#include<memory.h>#include<queue>using namespace std;char g[55][55];int dis[30][30];int j[15]; int tot;bool vis1[55][55];bool vis2[30];int step[55][55];int t, w, h, L, m;int dx[] = { -1, 1, 0, 0 };int dy[] = { 0, 0, -1, 1 };bool check(int x, int y){    if (vis1[x][y])return false;    if (x < 0 || x>h - 1 || y<0 || y>w - 1)return false;    if (g[x][y] == '*')return false;    return true;}void bfs(int x, int y, int tag){    memset(vis1, 0, sizeof(vis1));    memset(step, 0, sizeof(step));    queue<int>q;    q.push(x*w + y);    vis1[x][y] = 1;    step[x][y] = 0;    while (!q.empty()){        int cur = q.front();        q.pop();        int x = cur / w;         int y = cur%w;        for (int i = 0; i < 4; ++i){            int nx = x + dx[i];            int ny = y + dy[i];            if (check(nx, ny)){                vis1[nx][ny] = 1;                step[nx][ny] = step[x][y] + 1;                if (g[nx][ny] == '@')dis[tag][0] = step[nx][ny];                else if (g[nx][ny] == '<')dis[tag][m + 1] = step[nx][ny];                else if ('A' <= g[nx][ny] && g[nx][ny] <= 'J') dis[tag][g[nx][ny] - 'A' + 1] = step[nx][ny];                q.push(nx*w+ny);            }        }    }}int ans;void dfs(int tag,int val,int t){    if (t > L)return;    if (ans == tot)return;    if (tag > m&&val > ans){        ans = val;        return;    }    for (int i = 0; i <= m + 1; ++i){        if (!vis2[i] && dis[tag][i]){            vis2[i] = 1;            dfs(i, val + j[i], t + dis[tag][i]);            vis2[i] = 0;        }    }}int main(){    //freopen("in.txt", "r", stdin);    scanf("%d", &t);    for (int c = 1; c <= t; ++c){        memset(dis, 0, sizeof(dis));        tot = 0;        scanf("%d %d %d %d", &w, &h, &L, &m);        for (int i = 1; i <= m; ++i){            scanf("%d", &j[i]);            tot += j[i];        }        j[0] = 0; j[m + 1] = 0;//0代表入口，m+1代表出口，1~m代表宝石        for (int i = 0; i < h; ++i){            scanf("%s", g[i]);        }                for (int i = 0; i < h; ++i){            for (int j = 0; j < w; ++j){                if (g[i][j] == '@')bfs(i, j, 0);                else if (g[i][j] == '<')bfs(i, j, m + 1);                else if ('A' <= g[i][j] && g[i][j] <= 'J')bfs(i, j, g[i][j] - 'A' + 1);            }        }        ans = -1;        memset(vis2, 0, sizeof(vis2));        vis2[0] = 1;        dfs(0,0,0);        printf("Case %d:\n", c);        if (ans >= 0)printf("The best score is %d.\n", ans);        else printf("Impossible\n");        if (c <t)printf("\n");    }    return 0;}

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