HDU 4858 project management chart governance (Composite Algorithm)

HDU 4858 Project Management

Question: give n (<= 100000) vertices and m edges (<= N + 10). There may be duplicate edges. Each vertex has a value of Val, which is first recognized as 0.

Two operations.

Operation 1: Add addx to the value of vertex x.

Operation 2: output the Val sum of the neighboring points of X.

Analysis: simple optimization operations 1 or 2 won't work.

Method 1: vertices in the graph are divided into two types of vertices. For vertices whose degree is greater than SQRT (N), for vertices whose degree is less than or equal to SQRT (N), it is light. The number of key points is less than SQRT (n. Focuses and light points are handled separately.

Method 2: You can also classify neighboring points for each vertex. Vertices greater than this degree are classified into one type, which is equal to the degree of this point. vertices smaller than this point are classified into one type. Process them separately.

Method 1:

const int maxn = 100100;vector<int>v[maxn], vgt[maxn];int d[maxn];int val[maxn];int sum[maxn];int n, m;const int MM = 3333;int main (){    int T;    cin >> T;    while (T--)    {        scanf("%d%d", &n, &m);        memset(d, 0, sizeof(d));        memset(val, 0, sizeof(val));        memset(sum, 0, sizeof(sum));        for (int i = 1; i <= n; i++)        {            v[i].clear(); vgt[i].clear();        }        for (int i = 0; i < m; i++)        {            int x, y;            scanf("%d%d", &x, &y);            v[x].push_back(y); v[y].push_back(x);            d[x]++; d[y]++;        }        for (int i = 1; i <= n; i++)        {            for (int j = 0; j < v[i].size(); j++)            {                int r = v[i][j];                if (d[r] > MM) vgt[i].push_back(r);            }        }        int Q;        cin >> Q;        while (Q--)        {            int op;            int x, addx;            scanf("%d", &op);            if (!op)            {                scanf("%d%d", &x, &addx);                val[x] += addx;                if (d[x] <= MM)                {                    for (int i = 0; i < vgt[x].size(); i++)                    {                        int y = vgt[x][i];                        sum[y] += addx;                    }                }            }            else            {                scanf("%d", &x);                int ans = 0;                if (d[x] <= MM)                {                    for (int i = 0; i < v[x].size(); i++)                    {                        ans += val[v[x][i]];                    }                }                else                {                    ans += sum[x];                    for (int i = 0; i < vgt[x].size(); i++)                    {                        int y = vgt[x][i];                        ans += val[y];                    }                }                printf("%d\n", ans);            }        }    }    return 0;}/**43 21 21 360 1 150 3 41 11 30 2 331 23 21 21 260 1 151 10 2 151 11 3*/

Method 2:

const int maxn = 100100;vector<int>v[maxn], vgt[maxn], veq[maxn];int d[maxn];int val[maxn];int sum[maxn];int n, m;const int MM = 3333;int main (){    int T;    cin >> T;    while (T--)    {        scanf("%d%d", &n, &m);        memset(d, 0, sizeof(d));        memset(val, 0, sizeof(val));        memset(sum, 0, sizeof(sum));        for (int i = 1; i <= n; i++)        {            v[i].clear(); vgt[i].clear(); veq[i].clear();        }        for (int i = 0; i < m; i++)        {            int x, y;            scanf("%d%d", &x, &y);            v[x].push_back(y); v[y].push_back(x);            d[x]++; d[y]++;        }        for (int i = 1; i <= n; i++)        {            for (int j = 0; j < v[i].size(); j++)            {                int r = v[i][j];                if (d[r] > d[i]) vgt[i].push_back(r);                else if (d[r] == d[i]) veq[i].push_back(r);            }        }        int Q;        cin >> Q;        while (Q--)        {            int op;            int x, addx;            scanf("%d", &op);            if (!op)            {                scanf("%d%d", &x, &addx);                val[x] += addx;                for (int i = 0; i < vgt[x].size(); i++)                    sum[vgt[x][i]] += addx;            }            else            {                scanf("%d", &x);                int ans = sum[x];                for (int i = 0; i < vgt[x].size(); i++)                    ans += val[vgt[x][i]];                for (int i = 0; i < veq[x].size(); i++)                    ans += val[veq[x][i]];                printf("%d\n", ans);            }        }    }    return 0;}/**43 21 21 360 1 150 3 41 11 30 2 331 23 21 21 260 1 151 10 2 151 11 3*/