1295: [SCOI2009]最长距离

链接

思路

　　每次选两个点，spfa，到每个点需要搬多少石头，再枚举两个点，判断是否可以在搬得石头的个数小于t的情况下，走到，取最大值。

　　zz的我，spfa都不会写了。。。

代码

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>

using namespace std;

const int N = 35;
const int INF = 1e9;

char s[N][N];
int mp[N][N],dis[N][N],q[100100];
bool vis[N][N];
int dx[4] = {0,0,1,-1},dy[4] = {1,-1,0,0};
int n,m,t,L,R;

#define getid(x,y) (x-1)*n+y
#define getx(z) (z-1)/n+1
#define gety(z) ((z%m)==0)?m:(z%m)

void bfs(int a,int b) {
    for (int i=1; i<=n; ++i) 
        for (int j=1; j<=m; ++j) dis[i][j] = INF,vis[i][j] = false;
    dis[a][b] = mp[a][b];
    vis[a][b] = true;
    L = 1,R = 0;
    q[++R] = getid(a,b);
    while (L <= R) {
        int x = getx(q[L]),y = gety(q[L]);L++;//潜在bug不可以gety(q[L++]);
        for (int i=0; i<4; ++i) {
            int xx = x + dx[i],yy = y + dy[i];
            if (xx < 1 || xx > n || yy > m || yy < 1) continue;
            if (dis[xx][yy] <= dis[x][y] + mp[xx][yy]) continue;
            dis[xx][yy] = dis[x][y] + mp[xx][yy];
            q[++R] = getid(xx,yy);
            vis[xx][yy] = true;
        }
        vis[x][y] = false;
    }
}

int main() {
    cin >> n >> m >> t;
    for (int i=1; i<=n; ++i) {
        scanf("%s",s[i]+1);
    }
    for (int i=1; i<=n; ++i) 
        for (int j=1; j<=m; ++j) 
            mp[i][j] = s[i][j]=='1';
    double ans = 0;
    for (int a=1; a<=n; ++a) {
        for (int b=1; b<=m; ++b) {
            bfs(a,b);
            for (int c=1; c<=n; ++c) // 也是从1开始 
                for (int d=1; d<=m; ++d) // 
                    if (mp[a][b] + dis[c][d] <= t) 
                        ans = max(ans,sqrt(1.0*(a-c)*(a-c)+1.0*(b-d)*(b-d)));
        }
    }
    printf("%.6lf",ans);    
    return 0;
}