CF 1042 E. Vasya and Magic Matrix

E. Vasya and Magic Matrix

http://codeforces.com/contest/1042/problem/E

题意：

　　一个n*m的矩阵，每个位置有一个元素，给定一个起点，每次随机往一个小于这个点位置走，走过去的值为欧几里得距离的平方，求期望的值。

分析：

　　逆推期望。

　　将所有值取出，按元素大小排序，然后最小的就是0，往所有大于它的转移即可，复杂度n^2，见下方考试代码。

　　考虑每个点，从所有小于它的元素转移。排序后，维护前缀和，可以做到O(1)转移。

　　$f[i] = \sum\limits_{j=1,val[j]<val[i]}f[j] + (x_j - x_i)^2 + (y_j - y_i) ^ 2$

　　$f[i] =\sum\limits_{j=1,val[j]<val[i]} f[j] + x_j^2 - 2x_jx_i + x_i^2 + y_j^2 - 2y_jy_i + y_i ^ 2$

代码：

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<iostream>
 #include<cctype>
 #include<set>
 #include<vector>
 #include<queue>
 #include<map>
 #define fi(s) freopen(s,"r",stdin);
 #define fo(s) freopen(s,"w",stdout);
 using namespace std;
 typedef long long LL;

 inline int read() {
     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
     for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
 }

 const LL mod = ;
 const int N = ;

 struct Node {
     int x, y, val;
     bool zh;
     bool operator < (const Node &A) const {
         return val < A.val;
     }
 }A[N];
 LL f[N], cnt[N], sumx[N], sumy[N], sumx2[N], sumy2[N];

 LL ksm(LL a,LL b) {
     LL ans = ;
     while (b) {
         if (b & ) ans = 1ll * ans * a % mod;
         a = 1ll * a * a % mod;
         b >>= ;
     }
     return ans;
 }

 inline void add(LL &x,LL y) { (x += y) >= mod ? (x -= mod) : x; }
 inline void sub(LL &x,LL y) { (x -= y) <  ? (x += mod) : x; }

 void solve2(int n) {

     A[].val = -;
     for (int i=; i<=n; ++i) {
         if (A[i].val == A[i - ].val) cnt[i] = cnt[i - ];
         else cnt[i] = i - ;
         sumx[i] = (sumx[i - ] + A[i].x) % mod;
         sumy[i] = (sumy[i - ] + A[i].y) % mod;
         sumx2[i] = (sumx2[i - ] + 1ll * A[i].x * A[i].x % mod) % mod;
         sumy2[i] = (sumy2[i - ] + 1ll * A[i].y * A[i].y % mod) % mod;
     }

     LL sum = , tmp = ;
     for (int i=; i<=n; ++i) {
         LL x2 = sumx2[cnt[i]];
         LL y2 = sumy2[cnt[i]];
         LL z1 = 1ll * sumx[cnt[i]] *  % mod * A[i].x % mod;
         LL z2 = 1ll * sumy[cnt[i]] *  % mod * A[i].y % mod;
         LL h1 = 1ll * cnt[i] * A[i].x % mod * A[i].x % mod;
         LL h2 = 1ll * cnt[i] * A[i].y % mod * A[i].y % mod;

         add(f[i], x2); add(f[i], y2);
         sub(f[i], z1); sub(f[i], z2);
         add(f[i], h1); add(f[i], h2);
         add(f[i], sum);

         f[i] = 1ll * f[i] * ksm(cnt[i], mod - ) % mod;
         if (A[i].zh) {
             cout << f[i]; return ;
         }
         add(tmp, f[i]); // 只有小于的时候才转移！！！
         if (A[i].val < A[i + ].val) add(sum, tmp), tmp = ;
     }

 }

 int main() {
     int n = read(), m = read(), tot = ;
     for (int i=; i<=n; ++i)
         for (int j=; j<=m; ++j)
             A[++tot].x = i, A[tot].y = j, A[tot].val = read(), A[tot].zh = false;

     int x = read(), y = read(), z = (x - ) * m + y;
     A[z].zh = true;

     sort(A + , A + tot + );

     solve2(tot);
     return ;
 }

比赛时代码，记录调试历程。

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<iostream>
 #include<cctype>
 #include<set>
 #include<vector>
 #include<queue>
 #include<map>
 #define fi(s) freopen(s,"r",stdin);
 #define fo(s) freopen(s,"w",stdout);
 using namespace std;
 typedef long long LL;

 inline int read() {
     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
     for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
 }

 const LL mod = ;
 const int N = ;

 struct Node {
     int x, y, val;
     bool zh;
     bool operator < (const Node &A) const {
         return val < A.val;
     }
 }A[N];

 LL inv[N], deg[N], f[N];
 //double dp[N];

 LL ksm(LL a,LL b) {
     LL ans = ;
     while (b) {
         if (b & ) ans = 1ll * ans * a % mod;
         a = 1ll * a * a % mod;
         b >>= ;
     }
     return ans;
 }

 LL Calc(int i,int j) {
     return ((A[i].x - A[j].x) * (A[i].x - A[j].x) % mod + (A[i].y - A[j].y) * (A[i].y - A[j].y) % mod) % mod;
 }

 void solve1(int n) {
     for (int i=; i<=n; ++i) {
 //        cout << A[i].val << ": ";
         if (deg[i]) {
 //            dp[i] = dp[i] / (double)(deg[i]);
             f[i] = 1ll * ksm(deg[i], mod - ) * f[i] % mod;
         }
         if (A[i].zh) {
             cout << f[i]; return ;
         }
         for (int j=i+; j<=n; ++j)
             if (A[j].val > A[i].val) {
 //                dp[j] = dp[j] + dp[i] + Calc(i, j);
 //                cout << A[j].val << " " << Calc(i, j) <<"--";
                 f[j] = (f[j] + f[i] + Calc(i, j)) % mod;
                 deg[j] ++;
             }
 //        puts("");
     }
 }

 int cnt[N], sumx[N], sumy[N], sumx2[N], sumy2[N];

 inline void add(LL &x,LL y) { (x += y) >= mod ? (x -= mod) : x; }
 inline void sub(LL &x,LL y) { (x -= y) <  ? (x += mod) : x; }

 void solve2(int n) {

     A[].val = -;
     for (int i=; i<=n; ++i) {
         if (A[i].val == A[i - ].val) cnt[i] = cnt[i - ];
         else cnt[i] = i - ;
         sumx[i] = (sumx[i - ] + A[i].x) % mod;
         sumy[i] = (sumy[i - ] + A[i].y) % mod;
         sumx2[i] = (sumx2[i - ] + 1ll * A[i].x * A[i].x % mod) % mod;
         sumy2[i] = (sumy2[i - ] + 1ll * A[i].y * A[i].y % mod) % mod;
     }

     LL sum = , tmp = ;
     for (int i=; i<=n; ++i) {
         LL x2 = sumx2[cnt[i]];
         LL y2 = sumy2[cnt[i]];
         LL z1 = 1ll * sumx[cnt[i]] *  % mod * A[i].x % mod;
         LL z2 = 1ll * sumy[cnt[i]] *  % mod * A[i].y % mod;
         LL h1 = 1ll * cnt[i] * A[i].x % mod * A[i].x % mod;
         LL h2 = 1ll * cnt[i] * A[i].y % mod * A[i].y % mod;

         add(f[i], x2); add(f[i], y2);
         sub(f[i], z1); sub(f[i], z2);
         add(f[i], h1); add(f[i], h2);
         add(f[i], sum);

         f[i] = 1ll * f[i] * ksm(cnt[i], mod - ) % mod;
         if (A[i].zh) {
             cout << f[i]; return ;
         }
         add(tmp, f[i]); // 只有小于的时候才转移！！！
         if (A[i].val < A[i + ].val) add(sum, tmp), tmp = ;
     }

 }

 int main() {
     int n = read(), m = read(), tot = ;
     for (int i=; i<=n; ++i)
         for (int j=; j<=m; ++j)
             A[++tot].x = i, A[tot].y = j, A[tot].val = read(), A[tot].zh = false;

     int x = read(), y = read(), z = (x - ) * m + y;
     A[z].zh = true;

     sort(A + , A + tot + );

 //    if (tot <= 1000) {
 //        solve1(tot) ;return 0;
 //    }
     solve2(tot);
     return ;
 }