CCF 201409-4 最优配餐 传送门
这道题很水, 不过刚开始超时了. 有一个收获: 对一张图里分散的点进行BFS时, 不必每次都将其作为源点, 进行BFS. 而可以全部压入队列中, 进行BFS.
#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
int map[1005][1005], cnt[1005][1005];
int lenth[1005][1005], n; // 路长
long long ans = 0;
bool vis[1005][1005];
int move[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
queue<pair<int,int> > Q;
void bfs()
{
while (!Q.empty()) {
register int x = Q.front().first, y = Q.front().second;
Q.pop();
for (register int i = 0; i < 4; ++i) {
register int xx = x + move[i][0], yy = y + move[i][1];
if (vis[xx][yy]) continue;
if (map[xx][yy] == -1 || map[xx][yy] == -2) continue;
if (xx < 1 || xx > n || yy < 1 || yy > n) continue;
vis[xx][yy] = true;
if (lenth[xx][yy] < lenth[x][y] + 1) continue;
lenth[xx][yy] = lenth[x][y] + 1;
Q.push(make_pair(xx, yy));
}
}
}
int main()
{
ios::sync_with_stdio(false);
register int m, k, d;
cin >> n >> m >> k >> d;
memset(map, 0, sizeof(map));
memset(cnt, 0, sizeof(cnt));
memset(lenth, 0x3f, sizeof(lenth));
memset(vis, false, sizeof(vis));
for (register int i = 0, x, y; i < m; ++i) {
cin >> x >> y;
map[x][y] = -2; // 分店
lenth[x][y] = 0;
}
for (register int i = 0, x, y, c; i < k; ++i) {
cin >> x >> y >> c;
map[x][y] = 1;
cnt[x][y] += c; // 购物数量
}
for (register int i = 0, x, y; i < d; ++i) {
cin >> x >> y;
map[x][y] = -1;
}
for (register int i = 1; i <= n; ++i) {
for (register int j = 1; j <= n; ++j) {
if (map[i][j] == -2) {
Q.push(make_pair(i, j));
vis[i][j] = true;
}
}
}
bfs();
for (register int i = 1; i <= n; ++i) {
for (register int j = 1; j <= n; ++j) {
if (map[i][j] == 1) {
ans += cnt[i][j]*lenth[i][j];
}
}
}
cout << ans;
}