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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> HDU 4856 Tunnels(bfs+狀壓dp)

HDU 4856 Tunnels(bfs+狀壓dp)

編輯:C++入門知識

HDU 4856 Tunnels(bfs+狀壓dp)


題目大意:給你一個N*N的圖讓你到達所有的“.”點,“#”不能通過,有m組每組有一個入口,一個出口,入口可以傳送到出口,不知道經過m組的先後順序,讓你求出走過所有的“.”的最小時間。

思路:先bfs出來所有的m之間的最短距離,然後dp[j][i] 表示,在j狀態下開始第i步的最小路程,枚舉找到一個最小的dp[1<


Tunnels

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1139 Accepted Submission(s): 344


Problem Description Bob is travelling in Xi’an. He finds many secret tunnels beneath the city. In his eyes, the city is a grid. He can’t enter a grid with a barrier. In one minute, he can move into an adjacent grid with no barrier. Bob is full of curiosity and he wants to visit all of the secret tunnels beneath the city. To travel in a tunnel, he has to walk to the entrance of the tunnel and go out from the exit after a fabulous visit. He can choose where he starts and he will travel each of the tunnels once and only once. Now he wants to know, how long it will take him to visit all the tunnels (excluding the time when he is in the tunnels).
Input The input contains mutiple testcases. Please process till EOF.
For each testcase, the first line contains two integers N (1 ≤ N ≤ 15), the side length of the square map and M (1 ≤ M ≤ 15), the number of tunnels.
The map of the city is given in the next N lines. Each line contains exactly N characters. Barrier is represented by “#” and empty grid is represented by “.”.
Then M lines follow. Each line consists of four integers x1, y1, x2, y2, indicating there is a tunnel with entrence in (x1, y1) and exit in (x2, y2). It’s guaranteed that (x1, y1) and (x2, y2) in the map are both empty grid.
Output For each case, output a integer indicating the minimal time Bob will use in total to walk between tunnels.
If it is impossible for Bob to visit all the tunnels, output -1.
Sample Input
5 4
....#
...#.
.....
.....
.....
2 3 1 4
1 2 3 5
2 3 3 1
5 4 2 1

Sample Output
7
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

#define LL __int64


using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn = 20;

struct node
{
    int x0, y0;
    int x1, y1;
} f[maxn];

struct node1
{
    int x, y;
};

int cnt;

char str[maxn][maxn];
bool vis[maxn][maxn];
int dis[maxn][maxn];
int d[maxn][maxn];
int dx[] = {0, 1, -1, 0};
int dy[] = {1, 0, 0, -1};

int dp[1<<16][maxn];
int mp[maxn][maxn];
int n, m;

void spfa(int s, int t)
{
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
        {
            vis[i][j] = false;
            d[i][j] = INF;
        }
    }
    node1 tmp;
    tmp.x = s;
    tmp.y = t;
    vis[s][t] = true;
    queuefp;
    fp.push(tmp);
    d[s][t] = 0;
    while(!fp.empty())
    {
        tmp = fp.front();
        fp.pop();
        for(int i = 0; i < 4; i++)
        {
            int x = tmp.x+dx[i];
            int y = tmp.y+dy[i];
            if(x <= n && x >= 1 && y <= n && y >= 1 && mp[x][y])
            {
                if(vis[x][y]) continue;
                d[x][y] = d[tmp.x][tmp.y]+1;
                vis[x][y] = true;
                node1 tmx;
                tmx.x = x;
                tmx.y = y;
                fp.push(tmx);
            }
        }
    }
}

int main()
{
    while(~scanf("%d %d",&n, &m))
    {
        memset(mp, 0, sizeof(mp));
        for(int i = 1; i <= n; i ++)
        {
            scanf("%s", str[i]+1);
            for(int j = 1; j <= n; j++)
                if(str[i][j] == '.') mp[i][j] = 1;
        }
        for(int i = 1; i <= m; i++) scanf("%d %d %d %d",&f[i].x0, &f[i].y0, &f[i].x1, &f[i].y1);
        for(int i = 1; i <= m; i++)
        {
            spfa(f[i].x1, f[i].y1);
            for(int j = 1; j <= m; j++)
            {
                if(i == j)
                {
                    dis[i][j] = 0;
                    continue;
                }
                dis[i][j] = d[f[j].x0][f[j].y0];
            }
        }
        memset(dp, INF, sizeof(dp));
        for(int i = 1; i <= m; i++) dp[1<<(i-1)][i] = 0;
        for(int i = 2; i <= m; i++)
        {
            for(int j = 0; j < (1<

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