Problem:
Score : 500 points

Problem Statement

In Takahashi Kingdom, which once existed, there are N cities, and some pairs of cities are connected bidirectionally by roads.
The following are known about the road network:

People traveled between cities only through roads. It was possible to reach any city from any other city, via intermediate cities if necessary.

Different roads may have had different lengths, but all the lengths were positive integers.

Snuke the archeologist found a table with N rows and N columns, A, in the ruin of Takahashi Kingdom.
He thought that it represented the shortest distances between the cities along the roads in the kingdom.

Determine whether there exists a road network such that for each u and v, the integer A_{u, v} at the u-th row and v-th column of A is equal to the length of the shortest path from City u to City v.
If such a network exist, find the shortest possible total length of the roads.

Constraints

1 \leq N \leq 300

If i ≠ j, 1 \leq A_{i, j} = A_{j, i} \leq 10^9.

A_{i, i} = 0

Inputs

Input is given from Standard Input in the following format:

N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
...
A_{N, 1} A_{N, 2} ... A_{N, N}

Outputs

If there exists no network that satisfies the condition, print -1.
If it exists, print the shortest possible total length of the roads.

Sample Input 1

3
0 1 3
1 0 2
3 2 0

Sample Output 1

3

The network below satisfies the condition:

City 1 and City 2 is connected by a road of length 1.

City 2 and City 3 is connected by a road of length 2.

City 3 and City 1 is not connected by a road.

Sample Input 2

3
0 1 3
1 0 1
3 1 0

Sample Output 2

-1

As there is a path of length 1 from City 1 to City 2 and City 2 to City 3, there is a path of length 2 from City 1 to City 3.
However, according to the table, the shortest distance between City 1 and City 3 must be 3.

Thus, we conclude that there exists no network that satisfies the condition.

Sample Input 3

5
0 21 18 11 28
21 0 13 10 26
18 13 0 23 13
11 10 23 0 17
28 26 13 17 0

Sample Output 3

82

Sample Input 4

3
0 1000000000 1000000000
1000000000 0 1000000000
1000000000 1000000000 0

Sample Output 4

3000000000

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
import java.util.*;
import java.util.function.BiFunction;
import java.util.function.Consumer;
import java.util.function.Supplier;

class Edge {
    public int distance;
    public int from;
    public int to;
    public Edge(int distance, int from, int to) {
        this.distance = distance;
        this.from = from;
        this.to = to;
    }
}

public class Main {
    public static void main(String[] args) {
        final FastScanner fs = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);

        int N = fs.nextInt();

        int[][] A = new int[N][N];
        boolean[][] B = new boolean[N][N];
        for (int i = 0; i < N; i++) {
            Arrays.fill(B[i], true);
        }

        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                int distance = fs.nextInt();
                A[i][j] = distance;
            }
        }

        for (int k = 0; k < N; k++) {
            for (int i = 0; i < N; i++) {
                for (int j = 0; j < N; j++) {
                    int newMin = A[i][k] + A[k][j];
                    if (newMin < A[i][j]) {
                        System.out.println(-1);
                        return;
                    }

                    if (newMin == A[i][j] && i != k && k != j) {
                        B[i][j] = false;
                    }
                }
            }
        }

        long weight = 0;
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                if (B[i][j]) {
                    weight += A[i][j];
                }
            }
        }

        System.out.println(weight / 2);
    }

    static void debug(Object... os) {
        System.err.println(Arrays.deepToString(os));
    }
    // Execute func for all patterns can be represented with n bit.
    // O(n2^n)
    static void forAllBit(int n, Consumer<List<Boolean>> func) {
        for (int i = 0; i < (1 << n); i++) {
            List<Boolean> arg = new ArrayList();
            for (int j = 0; j < n; j++) {
                boolean isJthBitSet = ((i >> j) & 1) == 1;
                arg.add(isJthBitSet);
            }
            func.accept(arg);
        }
    }

    static class Pair<F extends Comparable<F>, S extends Comparable<S>> implements Comparable<Pair<F, S>> {
        F f;
        S s;

        Pair() {
        }

        Pair(F f, S s) {
            this.f = f;
            this.s = s;
        }

        Pair(Pair<F, S> p) {
            f = p.f;
            s = p.s;
        }

        @Override
        public int compareTo(Pair<F, S> p) {
            if (f.compareTo(p.f) != 0) {
                return f.compareTo(p.f);
            }
            return s.compareTo(p.s);
        }

        @Override
        public int hashCode() {
            return f.hashCode() ^ s.hashCode();
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) {
                return true;
            }
            if (o == null || this.f == null || this.s == null) {
                return false;
            }
            if (this.getClass() != o.getClass()) {
                return false;
            }
            Pair p = (Pair) o;
            return this.f.equals(p.f) && this.s.equals(p.s);
        }

        @Override
        public String toString() {
            return "{" + f.toString() + ", " + s.toString() + "}";
        }
    }


    public static int nCk(int n, int k, int M) {
        long ret = 1;
        int min = Math.min(k, n - k);
        for (int i = 1; i <= min; i++) {
            ret = (ret * pow(i, M - 2, M)) % M;
        }
        for (int i = n - min + 1; i <= n; i++) {
            ret = (ret * i) % M;
        }
        return (int) ret;
    }

    //return a^b mod M O(logB)
    public static long pow(long a, long b, int M) {
        long ret = 1;
        long tmp = a;
        while (b > 0) {
            if ((b & 1) == 1) ret = (ret * tmp) % M;
            tmp = (tmp * tmp) % M;
            b = b >> 1;
        }
        return ret;
    }


    public static int find(int[] tree, int idx) {
        if (tree[idx] == idx) return idx;
        else return tree[idx] = find(tree, tree[idx]);
    }

    public static void union(int[] tree, int idx1, int idx2) {
        int root1 = find(tree, idx1);
        int root2 = find(tree, idx2);
        tree[root2] = root1;
    }

    public static int lowerBound(int[] list, int target) {
        int ok = list.length;
        int ng = -1;
        while (Math.abs(ok - ng) > 1) {
            int mid = (ok + ng) / 2;
            if (list[mid] >= target) {
                ok = mid;
            } else {
                ng = mid;
            }
        }
        return ok;
    }

    public static int lowerBound(long[] list, long target) {
        int ok = list.length;
        int ng = -1;
        while (Math.abs(ok - ng) > 1) {
            int mid = (ok + ng) / 2;
            if (list[mid] >= target) {
                ok = mid;
            } else {
                ng = mid;
            }
        }
        return ok;
    }

    public static <T extends Comparable<? super T>> int lowerBound(List<T> list, T target) {
        return ~Collections.binarySearch(list, target, (x, y) -> (x.compareTo(y) >= 0) ? 1 : -1);
    }

    public static <T extends Comparable<? super T>> int upperBound(List<T> list, T target) {
        return ~Collections.binarySearch(list, target, (x, y) -> (x.compareTo(y) > 0) ? 1 : -1);
    }
}


class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        } else {
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }

    private int readByte() {
        if (hasNextByte()) return buffer[ptr++];
        else return -1;
    }

    private boolean isPrintableChar(int c) {
        return 33 <= c && c <= 126;
    }

    public boolean hasNext() {
        while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
        return hasNextByte();
    }

    public String next() {
        if (!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while (isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    public long nextLong() {
        if (!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while (true) {
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            } else if (b == -1 || !isPrintableChar(b)) {
                return minus ? -n : n;
            } else {
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }

    public int nextInt() {
        long nl = nextLong();
        if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
        return (int) nl;
    }

    public double nextDouble() {
        return Double.parseDouble(next());
    }
}

class BIT<T> {
    int n;
    ArrayList<T> bit;
    BiFunction<T, T, T> bif;

    /**
     * 1-indexed なBinary Indexed Treeを構築する
     *
     * @param n   容量
     * @param bif 適用させる関数
     * @param sup 初期値
     */
    BIT(int n, BiFunction<T, T, T> bif, Supplier<T> sup) {
        this.n = n;
        bit = new ArrayList<>(n + 1);
        for (int i = 0; i < n + 1; ++i) {
            bit.add(sup.get());
        }
        this.bif = bif;
    }

    /**
     * iの位置の値をvで更新する
     *
     * @param i index
     * @param v 新しい値
     */
    void set(int i, T v) {
        for (int x = i; x <= n; x += x & -x) {
            bit.set(x, bif.apply(bit.get(x), v));
        }
    }

    /**
     * クエリー
     *
     * @param defaultValue 初期値
     * @param i            index
     * @return [1, i]までfを適用した結果
     */
    T reduce(T defaultValue, int i) {
        T ret = defaultValue;
        for (int x = i; x > 0; x -= x & -x) {
            ret = bif.apply(ret, bit.get(x));
        }
        return ret;
    }
}

class SegmentTree<T> {
    int n;
    ArrayList<T> dat;
    BiFunction<T, T, T> bif;
    Supplier<T> sup;

    /**
     * 0-indexed なSegment Treeを構築する
     *
     * @param n_  要求容量
     * @param bif 適用させる関数
     * @param sup 初期値
     */
    SegmentTree(int n_, BiFunction<T, T, T> bif, Supplier<T> sup) {
        n = 1;
        while (n < n_) n *= 2;

        dat = new ArrayList<>(2 * n - 1);
        for (int i = 0; i < 2 * n - 1; ++i) {
            dat.add(sup.get());
        }
        this.bif = bif;
        this.sup = sup;
    }

    /**
     * kの位置の値をvで更新する
     *
     * @param k index
     * @param v 新しい値
     */
    void set(int k, T v) {
        k += n - 1;
        dat.set(k, v);
        while (k > 0) {
            k = (k - 1) / 2;
            dat.set(k, bif.apply(dat.get(k * 2 + 1), dat.get(k * 2 + 2)));
        }
    }

    /**
     * クエリー
     *
     * @param l はじめ
     * @param r おわり
     * @return [l, r)での演算bifを適用した結果を返す
     */
    T reduce(int l, int r) {
        return _reduce(l, r, 0, 0, n);
    }

    T _reduce(int a, int b, int k, int l, int r) {
        if (r <= a || b <= l) return sup.get();
        if (a <= l && r <= b) return dat.get(k);
        T vl = _reduce(a, b, k * 2 + 1, l, (l + r) / 2);
        T vr = _reduce(a, b, k * 2 + 2, (l + r) / 2, r);
        return bif.apply(vl, vr);
    }
}

class UnionFind {
    int[] par;

    UnionFind(int n) {
        par = new int[n];
        for (int i = 0; i < n; ++i) {
            par[i] = i;
        }
    }

    int find(int x) {
        if (par[x] == x) {
            return x;
        }
        return par[x] = find(par[x]);
    }

    boolean same(int x, int y) {
        return find(x) == find(y);
    }

    void union(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) {
            return;
        }
        par[x] = y;
    }
}