HDU 3681 Prison Break (二分 + bfs + TSP)

题意：给定上一个 n * m的矩阵，你的出发点是 F，你初始有一个电量，每走一步就会少1，如果遇到G，那么就会加满，每个G只能第一次使用，问你把所有的Y都经过，初始电量最少是多少。

析：首先先预处理每个F，G，Y的最短距离，用 bfs 可以实现，然后再二分电量，进行判断，在进行判断时，dp[s][i] 表示已经走点的状态是s，当前在 i，然后如果能走过所有的就就是true，否则是false。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
//#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 15 + 10;
const int maxm = 1e5 + 10;
const int mod = 50007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

char s[maxn][maxn];
vector<P> v;
int dp[(1<<16)+10][20];
int dist[20][20];
int vis[20][20];

int bfs(int ss, int tt){
  queue<P> q;  ms(vis, -1);
  vis[v[ss].fi][v[ss].se] = 0;
  q.push(v[ss]);

  while(!q.empty()){
    P p = q.front();  q.pop();
    for(int i = 0; i < 4; ++i){
      int x = p.fi + dr[i];
      int y = p.se + dc[i];
      if(!is_in(x, y) || vis[x][y] != -1 || s[x][y] == 'D')  continue;
      vis[x][y] = vis[p.fi][p.se] + 1;
      P qq = P(x, y);
      if(qq == v[tt])  return vis[x][y];
      q.push(qq);
    }
  }
  return -1;
}

int start, st;;
bool judge(int mid){
  ms(dp, -1);
  dp[1][0] = mid;
  int all = 1<<v.sz;
  for(int i = 0; i < all; ++i)
    for(int j = 0; j < v.sz; ++j){
      if(dp[i][j] >= 0 && (i & st) == st)  return true;
      if(dp[i][j] <= 0)  continue;
      for(int k = 0; k < v.sz; ++k){
        if(dist[j][k] == -1 || i&1<<k)  continue;
        int &ans = dp[i|1<<k][k];
        ans = max(ans, dp[i][j] - dist[j][k]);
        if(ans >= 0 && s[v[k].fi][v[k].se] == 'G')  ans = mid;
        if(ans >= 0 && ((i|1<<k)&st) == st)  return true;
      }
    }
  return false;
}

int main(){
  while(scanf("%d %d", &n, &m) == 2 && n+m){
    v.clear(); st = 0;
    for(int i = 0; i < n; ++i){
      scanf("%s", s[i]);
      for(int j = 0; j < m; ++j){
        if(s[i][j] == 'D' || s[i][j] == 'S')  continue;
        if(s[i][j] == 'F')  start = v.sz;
        else if(s[i][j] == 'Y')  st |= 1<<v.sz;
        v.push_back(P(i, j));
      }
    }
    if(s[v[0].fi][v[0].se] == 'Y')  st ^= 1|1<<start;
    swap(v[start], v[0]);
    for(int i = 0; i < v.sz; ++i)
      for(int j = i+1; j < v.sz; ++j)
        dist[i][j] = dist[j][i] = bfs(i, j);
    int l = 0, r = n * m;
    if(!judge(r)){ puts("-1");  continue; }
    while(l <= r){
      int m = l + r >> 1;
      if(judge(m))  r = m - 1;
      else l = m + 1;
    }
    printf("%d
", l);
  }
  return 0;
}