USACO2013 island travels

题意：一个R行C列的矩阵，'X'表示地，'S'表示浅水，'.'表示不能走的深水。连通的X视为一个岛（不超过15个）。现在要走完所有岛，求最少的踩在浅水格子的次数。

题解：岛屿不超过15个，明显的暗示可以用状态压缩DP跑旅行商问题。但是这题需要较多的预处理。首先给每个X连通块标上岛屿的序号，然后对每一个岛屿，将它直接相邻的浅水格子压入队列跑BFS即可求出所有岛屿到他的距离。然后记得一定要跑一次Floyd！！DP的状态定义为f[s, i]表示经过了的岛屿的集合为s，最后到达i的最少路径长。

考试的时候犯了很SB的两个错误，说明写代码的时候思路不清晰。以后最好写一个函数就检查一个。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
#define Min(a,b) ((a)<(b)?(a):(b))
#define inmap(x,y) (x>0 && y>0 && x<=R && y<=C)
const int MAXN = 105;
const int dd[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
int N, R, C;
int dis[20][20];
char sea[MAXN][MAXN];
int cnt;
const int INF = 0x3f3f3f3f;#define code(x,y) ((x-1)*C+y)
#define getx(s) ((s-1)/C+1)
#define gety(s) ((s-1)%C+1)
void mark(int x, int y)
{queue<int> Q;Q.push(code(x,y));++cnt;while (!Q.empty()) {x = getx(Q.front());y = gety(Q.front());sea[x][y] = cnt;Q.pop();for (int i = 0; i<4; ++i) {int tx = x+dd[i][0], ty = y+dd[i][1];if (inmap(tx,ty) && sea[tx][ty]=='X')Q.push(code(tx, ty));}}
}int step[MAXN][MAXN];
void BFS(int id)
{int i, j, k, x, y, tx, ty;queue<int> Q;memset(step, 0, sizeof step);for (i = 1; i<=R; ++i)for (j = 1; j<=C; ++j)if (sea[i][j] == id)for (k = 0; k<4; ++k) {tx = i+dd[k][0], ty = j+dd[k][1];if (inmap(tx,ty) && sea[tx][ty]=='S')//考试的时候居然忘记判断是否是浅水就直接入队了！！Q.push(code(tx, ty)), step[tx][ty] = 1;}while (!Q.empty()) {x = getx(Q.front()), y = gety(Q.front());Q.pop();for (i = 0; i<4; ++i) {tx = x+dd[i][0], ty = y+dd[i][1];if (!inmap(tx,ty)) continue;if (sea[tx][ty]=='S' && !step[tx][ty])Q.push(code(tx,ty)), step[tx][ty] = step[x][y]+1;else if (1<=sea[tx][ty] && sea[tx][ty]<=cnt && sea[tx][ty]!=id) {j = sea[tx][ty];dis[id][j] = Min(dis[id][j], step[x][y]),dis[j][id] = dis[id][j];}}}
}void Floyd()//考试的时候忘记加floyd了，以后一定要注意这些细节。
{for (int k = 1; k<=cnt; ++k)for (int i = 1; i<=cnt; ++i)for (int j = 1; j<=cnt; ++j)dis[i][j] = Min(dis[i][j], dis[i][k] + dis[k][j]);
}int f[(1<<16)+1][16];
int ans = INF;
void DoDP()
{ memset(f, 0x3f, sizeof f);int s, s1, i, j;for (i = 1; i<=cnt; ++i)f[1<<(i-1)][i] = 0;for (s = 1; s < (1<<cnt); ++s)for (i = 1; i<=cnt; ++i){if (!(s&(1<<(i-1)))) continue;s1 = s ^ (1<<(i-1));for (j = 1; j<=cnt; ++j){if (!(s1&(1<<(j-1)))) continue;f[s][i] = Min(f[s][i], f[s1][j]+dis[j][i]);}}for (i = 1; i<=cnt; ++i)ans = Min(ans, f[(1<<cnt)-1][i]);
}int main()
{ int i, j; scanf("%d%d", &R, &C); memset(dis, 0x3f, sizeof dis); for (i = 1; i<=R; ++i) scanf("%s", sea[i]+1); for (i = 1; i<=R; ++i) for (j = 1; j<=C; ++j) if (sea[i][j] == 'X') mark(i, j); for (i = 1; i<=cnt; ++i) BFS(i); Floyd();DoDP();printf("%d\n", ans);return 0;
}

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