hdu 3682

来源：互联网发布：c语言九九乘法编程编辑：程序博客网时间：2024/06/15 05:27
二分 + 状压dp
dp[i][j]表示终点为i，访问状态为j剩余的最大能量
二分一个上限能量
dp[i][j] 可以由dp[k][s]转移过来，表示从k点走到i点，如果k点是一个电池的话那么只要满足dp[k][s]剩余的能量 >= 0，否则则是dp[k][s] - dist[i][k]
#include <cstdio>#include <cstring>#include <queue>using namespace std;const int maxn = 30;const int INF = 10007;struct Point {    int x, y, d;};int tot;struct Point point[maxn];int type[maxn], num[maxn][maxn];int dp[17][1 << (16)];int goal;int n, m;int step[4][2] = {{0, 1}, {0,-1},{-1, 0}, {1, 0}};char maps[maxn][maxn];queue<struct Point> que;int dist[maxn][maxn], vis[maxn][maxn];int pre[17][1 << 16];void bfs(int source) {    memset(vis, 0, sizeof(vis));    for (int i = 0; i <= tot; i++) {        dist[source][i] = INF;    }    memset(vis, 0, sizeof(vis));    dist[source][source] = 0;    point[source].d = 0;    que.push(point[source]);    vis[point[source].x][point[source].y] = 1;    while (!que.empty()) {        struct Point p = que.front();        que.pop();        for (int k = 0; k < 4; k++) {            int next_x = p.x + step[k][0];            int next_y = p.y + step[k][1];            if (next_x >= 0 && next_x < n && next_y >= 0 && next_y < m && !vis[next_x][next_y] && maps[next_x][next_y] != 'D') {                vis[next_x][next_y] = 1;                struct Point a;                a.x = next_x;                a.y = next_y;                a.d = p.d + 1;                if (num[next_x][next_y] != -1) {                    dist[source][num[next_x][next_y]] = p.d + 1;                }                que.push(a);            }        }    }}bool check(int p) {    memset(dp, -1, sizeof(dp));    dp[0][1] = p;    for (int i = 1; i < (1 << tot); i++) {        for (int j = 0; j < tot; j++) {            if (!(i & (1 << j))) {                continue;            }            for (int k = 0; k < tot; k++) {                if (i & (1 << k) && k != j) {                    if (type[k] == 1) {                        if (dp[k][i ^ (1 << j)] < 0) {                            continue;                        }                        dp[j][i] = max(dp[j][i], p - dist[k][j]);                    }                    else {                        dp[j][i] = max(dp[j][i], dp[k][i ^ (1 << j)] - dist[k][j]);                    }                }            }            if (((i & goal) == goal) && dp[j][i] >= 0) {                return true;            }        }    }    return false;}int main() {//    freopen("in.txt", "r", stdin);    while (scanf("%d%d", &n, &m) != EOF) {        if (!n && !m) {            break;        }        for (int i = 0; i < n; i++) {            scanf("%s", maps[i]);        }        tot = 1;        goal = 0;        for (int i = 0; i < n; i++) {            for (int j = 0; j < m; j++) {                num[i][j] = -1;                if (maps[i][j] == 'F') {                    point[0].x = i;                    point[0].y = j;                    num[i][j] = 0;                }                else if (maps[i][j] == 'G') {                    point[tot].x = i;                    point[tot].y = j;                    num[i][j] = tot;                    type[tot] = 1;                    tot++;                }                else if (maps[i][j] == 'Y') {                    point[tot].x = i;                    point[tot].y = j;                    type[tot] = 2;                    num[i][j] = tot;                    goal |= (1 << tot);                    tot++;                }            }        }        for (int i = 0; i < tot; i++) {            bfs(i);        }        int flag = 0;        for (int i = 1; i < tot; i++) {            if (type[i] == 2 && dist[0][i] == INF) {                flag = 1;                break;            }        }        if (flag) {            puts("-1");            continue;        }        int low = 0;        int high = INF;        while (low <= high) {            int mid = (low + high) >> 1;            if (check(mid)) {                high = mid - 1;            }            else {                low = mid + 1;            }        }        printf("%d\n", low);    }    return 0;}
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