HDU - 4966 GGS-DDU (最小树形图)

题目大意：有一个人，想学习N个科目，每个科目都有相应的层次

有M个课程，M个课程的要求是，你的第c个科目的层次要达到l1，才可以参加，参加完这个课程后，你需要缴费money，但你的第d个科目的层次会达到l2

现在问，如何花最少的钱，使得每个科目的层次都达到最高

解题思路：每个科目的每个层次都看成一个点，每个科目的层次0和根连边，费用0

每个科目的高层次向下一个低层次连边，费用为0(因为你高层次的会了就表示低层次的也会了)

接着按要求连边就可以了

预处理还是挺麻烦的

[code]#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int MAXNODE = 1010;
const int MAXEDGE = 1000100;
typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
    int u, v;
    Type dis;
    Edge() {}
    Edge(int u, int v, Type dis): u(u), v(v), dis(dis) {}
};

struct Directed_MT{
    int n, m;
    Edge edges[MAXEDGE];
    int vis[MAXNODE];
    int pre[MAXNODE];
    int id[MAXNODE];
    Type in[MAXNODE];

    void init(int n) {
        this->n = n;
        m = 0;
    }

    void AddEdge(int u, int v, Type dis) {
        edges[m++] = Edge(u, v, dis);
    }

    Type DirMt(int root) {
        Type ans = 0;
        while (1) {
            //初始化
            for (int i = 0; i < n; i++) in[i] = INF;

            for (int i = 0; i < m; i++) {
                int u = edges[i].u;
                int v = edges[i].v;
                //找寻最小入边，删除自环
                if (edges[i].dis < in[v] && u != v) {
                    in[v] = edges[i].dis;
                    pre[v] = u;
                }
            }

            for (int i = 0; i < n; i++) {
                if (i == root) continue;
                //如果没有最小入边，表示该点不连通，则最小树形图形成失败
                if (in[i] == INF) return -1;
            }

            int cnt = 0;//记录缩点
            memset(id, -1, sizeof(id));
            memset(vis, -1, sizeof(vis));
            in[root] = 0;//树根不能有入边
            for (int i = 0; i < n; i++) {
                ans += in[i];
                int v = i;
                //找寻自环
                while (vis[v] != i && id[v] == -1 && v != root) {
                    vis[v] = i;
                    v = pre[v];
                }
                //找到自环
                if (v != root && id[v] == -1) {
                    for (int u = pre[v]; u != v; u = pre[u]) 
                        id[u] = cnt;
                    id[v] = cnt++;
                }
            }

            //如果没有自环了，表示最小树形图形成成功了
            if (cnt == 0) break;

            //找到那些不是自环的，重新给那些点进行标记
            for (int i = 0; i < n; i++) 
                if (id[i] == -1) id[i] = cnt++;

            for (int i = 0; i < m; i++) {
                int v = edges[i].v;
                edges[i].v = id[edges[i].v];
                edges[i].u = id[edges[i].u];
                if (edges[i].u != edges[i].v) 
                    edges[i].dis -= in[v];
            }
            //缩点完后，点的数量就边了
            n = cnt;
            root = id[root];
        }
        return ans;
    }
}MT;

const int N = 60;
const int M = 600;
int n, m;
int level
, Sum
;
int dis[M][M];

void init() {
    Sum[0] = 0;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &level[i]);
        level[i]++;
        if (i == 1) Sum[i] = level[i];
        else Sum[i] = Sum[i - 1] + level[i];
    }

    Sum[n + 1] = Sum
 + 1;
    MT.init(Sum[n + 1]);
    //跟根结点连边
    for (int i = 1; i <= n; i++)
        if (i == 1) MT.AddEdge(0, i, 0);
        else MT.AddEdge(0, Sum[i - 1] + 1, 0);

    //高层次的连向低层次的
    for (int i = 1; i <= n; i++) 
        for (int j = level[i] - 1; j > 0; j--) 
            MT.AddEdge(j + 1 + Sum[i - 1], j + Sum[i - 1], 0);

    memset(dis, 0x3f, sizeof(dis));
    int c, l1, d, l2, money;
    //课程
    for (int i = 1; i <= m; i++) {
        scanf("%d%d%d%d%d", &c, &l1, &d, &l2, &money);
        for (int u = Sum[c - 1] + 1 + l1; u <= Sum[c]; u++) {
            int v = Sum[d - 1] + l2 + 1;
            dis[u][v] = min(dis[u][v], money);
        }
    }

    //取课程的最小费用，连边
    for (int i = 1; i < Sum[n + 1]; i++)
        for (int j = 1; j < Sum[n + 1]; j++)
            if (dis[i][j] != INF)
                MT.AddEdge(i, j, dis[i][j]);
    printf("%d\n", MT.DirMt(0));
}

int main() {
    while (scanf("%d%d", &n, &m) != EOF && n + m) {
        init();
    }
    return 0;
}