HDU 4966 GGS-DDU

来源：互联网发布：天刀霸气真武捏脸数据编辑：程序博客网时间：2024/05/14 10:02

题意：

你有n个课程每个课程有一个规定的毕业学分修学分有m种方式每种方式要求先修到x课程x'学分以上才能花费money去修y课程并且将学分修到y' 问最少花费多少可以毕业

思路：

一开始想费用流建完图发现一个问题解决不掉那就是一条边如果流过多次怎样才能让费用只计算一次所以换思路

我们知道为了应付“ 学分修到y' ”这个条件高层学分一定要“覆盖”低层学分那么就想到以每个学分为一个点高层向低层连边费用为0 同时修学分的方式是输入的直接建边费用为输入费用一开始只能走到0学分所以超级源点向所有0学分连边费用0 那么此时我们想要的是从源点花最小费用走到所有点这就是有向图最小生成树即最小树形图

代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<queue>using namespace std;#define M 610#define inf 2000000000int n, m, ans, tot, S;int head[M], id[M], in[M], pre[M], vis[M], node[M][M], s[M];struct ed {    int u, v, w, next;} ed[M * M];int ZL(int root, int V, int E) {    int ret = 0;    for (;;) {        for (int i = 0; i < V; i++) {            id[i] = -1;            vis[i] = -1;            in[i] = inf;        }        for (int i = 0; i < E; i++) {            int u = ed[i].u;            int v = ed[i].v;            if (ed[i].w < in[v] && u != v) {                pre[v] = u;                in[v] = ed[i].w;            }        }        for (int i = 0; i < V; i++) {            if (i == root)                continue;            if (in[i] == inf)                return -1;        }        int cnt = 0;        in[root] = 0;        for (int i = 0; i < V; i++) {            ret += in[i];            int v = i;            while (vis[v] != i && id[v] == -1 && v != root) {                vis[v] = i;                v = pre[v];            }            if (v != root && id[v] == -1) {                for (int u = pre[v]; u != v; u = pre[u])                    id[u] = cnt;                id[v] = cnt++;            }        }        if (cnt == 0)            break;        for (int i = 0; i < V; i++)            if (id[i] == -1)                id[i] = cnt++;        for (int i = 0; i < E; i++) {            int u = ed[i].u;            int v = ed[i].v;            ed[i].u = id[u];            ed[i].v = id[v];            if (id[u] != id[v])                ed[i].w -= in[v];        }        V = cnt;        root = id[root];    }    return ret;}void add(int U, int V, int W) {    ed[tot].u = U;    ed[tot].v = V;    ed[tot].w = W;    ed[tot].next = head[U];    head[U] = tot++;}int main() {    int i, j, k, u, v, tmp;    while (scanf("%d%d", &n, &m) != EOF) {        if (!n && !m)            break;        k = 0;        for (i = 1; i <= n; i++) {            scanf("%d", &s[i]);            for (j = 0; j <= s[i]; j++)                node[i][j] = k++;        }        S = k++;        tot = 0;        memset(head, -1, sizeof(head));        for (i = 1; i <= n; i++) {            add(S, node[i][0], 0);            for (j = 1; j <= s[i]; j++)                add(node[i][j], node[i][j - 1], 0);        }        for (i = 1; i <= m; i++) {            scanf("%d%d", &u, &v);            tmp = node[u][v];            scanf("%d%d", &u, &v);            v = node[u][v];            u = tmp;            scanf("%d", &tmp);            add(u, v, tmp);        }        ans = ZL(S, k, tot);        printf("%d\n", ans);    }    return 0;}

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