标签：const vis int 搜题 几道 include bx dis
广搜好难/kk
P1162 填涂颜色
在原矩阵外再围一层 \(2\)，方便能够从边界搜索。
把输入数据中的 \(0\) 全都换成 \(2\)，然后再处理在封闭圈外的 \(2\)，将其变成 \(0\)。
#include <queue>
#include <cmath>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int A = 111;
const int dx[4] = {0, 0, 1, -1};
const int dy[4] = {1, -1, 0, 0};
inline int read() {
char c = getchar(); int x = 0, f = 1;
for ( ; !isdigit(c); c = getchar()) if (c == '-') f = -1;
for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
return x * f;
}
int n, a[A][A], vis[A][A];
struct node { int x, y; };
void bfs() {
queue <node> Q;
Q.push((node){0, 0});
a[0][0] = 0, vis[0][0] = 1;
while (!Q.empty()) {
int x = Q.front().x, y = Q.front().y;
Q.pop();
for (int i = 0; i < 4; i++) {
int bx = x + dx[i], by = y + dy[i];
if (bx >= 0 && bx <= n + 1 && by >= 0 && by <= n + 1 && !vis[bx][by] && a[bx][by] != 1) {
if (a[bx][by] == 2) {
vis[bx][by] = 1, a[bx][by] = 0;
Q.push((node){bx, by});
}
}
}
}
return;
}
int main() {
n = read();
for (int i = 0; i <= n + 1; i++) a[0][i] = 2;
for (int i = 0; i <= n + 1; i++) a[n + 1][i] = 2;
for (int i = 1; i <= n; i++) {
a[i][0] = a[i][n + 1] = 2;
for (int j = 1; j <= n; j++) {
a[i][j] = read();
if (a[i][j] == 0) a[i][j] = 2;
}
}
bfs();
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
cout << a[i][j] << " ";
}
puts("");
}
return 0;
}
P1443 马的遍历
广搜标记步数即可，像是个变异的最短路（大雾）。
注意厂宽。
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int A = 4e2 + 11;
const int inf = 0x3f3f3f3f;
const int dx[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
const int dy[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
inline int read() {
char c = getchar(); int x = 0, f = 1;
for ( ; !isdigit(c); c = getchar()) if (c == '-') f = -1;
for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
return x * f;
}
struct node { int x, y; };
int n, m, sx, sy, vis[A][A], dis[A][A];
int main() {
n = read(), m = read(), sx = read(), sy = read();
queue <node> Q;
memset(dis, inf, sizeof(dis));
Q.push((node){sx, sy});
dis[sx][sy] = 0, vis[sx][sy] = 1;
while (!Q.empty()) {
int x = Q.front().x, y = Q.front().y;
Q.pop(), vis[x][y] = 0;
for (int i = 0; i < 8; i++) {
int bx = x + dx[i], by = y + dy[i];
if (bx < 1 || bx > n || by < 1 || by > m) continue;
if (dis[x][y] + 1 < dis[bx][by]) {
dis[bx][by] = dis[x][y] + 1;
if (!vis[bx][by]) vis[bx][by] = 1, Q.push((node){bx, by});
}
}
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
printf("%-5d", dis[i][j] == inf ? -1 : dis[i][j]);
}
puts("");
}
return 0;
}
P3956 棋盘
直接跑最短路，广搜有点麻烦/kk。
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int A = 1e3 + 11;
const int inf = 0x3f3f3f3f;
inline int read() {
char c = getchar(); int x = 0, f = 1;
for ( ; !isdigit(c); c = getchar()) if (c == '-') f = -1;
for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
return x * f;
}
int flag, s, t, n, m;
int vis[A], e[A][A], x[A], y[A], w[A], dis[A];
struct node {
int x, y;
bool operator < (const node &b) const {
return y > b.y;
}
};
priority_queue <node> Q;
inline void Dij() {
memset(dis, inf, sizeof(dis));
Q.push((node){s, 0});
dis[s] = 0, vis[s] = 1;
while (!Q.empty()) {
int x = Q.top().x; Q.pop(), vis[x] = 0;
for (int i = 1; i <= m; i++)
if (dis[x] + e[x][i] < dis[i]) {
dis[i] = e[x][i] + dis[x];
if (!vis[i]) vis[i] = 1, Q.push((node){i, dis[i]});
}
}
}
int main() {
n = read(), m = read();
for (int i = 1; i <= m; i++) {
x[i] = read(), y[i] = read(), w[i] = read();
if (x[i] == 1 && y[i] == 1) s = i;
if (x[i] == n && y[i] == n) flag = 1, t = i;
}
if (!flag) x[m + 1] = n, y[m + 1] = n, t = m + 1;
memset(e, inf, sizeof(e));
for (int i = 1; i <= m; i++) {
for (int j = i + 1; j <= m; j++) {
if (abs(x[i] - x[j]) + abs(y[i] - y[j]) == 1)
e[i][j] = e[j][i] = abs(w[i] - w[j]);
if (abs(x[i] - x[j]) + abs(y[i] - y[j]) == 2)
e[i][j] = e[j][i] = 2 + abs(w[i] - w[j]);
}
}
if (!flag) {
for (int i = 1; i <= m; i++)
if (abs(x[i] - x[t]) + abs(y[i] - y[t]) == 1)
e[i][t] = e[t][i] = 2;
m++;
}
Dij();
printf("%d\n", dis[t] == inf ? -1 : dis[t]);
return 0;
}
P1032 字串变换
P1126 机器人搬重物
标签：const,vis,int,搜题,几道,include,bx,dis 来源： https://www.cnblogs.com/loceaner/p/13648736.html
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