hdu 3572 Task Schedule ISAP 网络流 构图

在此 dinic 也过时了 

从Lost那里抄过来的模板 复杂度是O(M*(N^2));

而dinic的复杂度也是O(M*(N^2)); 

理论复杂度虽然一样, 但实际却差很多, 这归根于Gap优化, 即相邻顶点标号差严格等于1, 若存在某个顶点标号的顶点数为0,

则说明不存在增广路; 

#include <iostream>
using namespace std; const int MAXN=1010; const int MAXM=500010; typedef struct {int v,next,val;} edge;

edge e[MAXM]; int p[MAXN],eid, ans; int n,m; int h[MAXN]; int gap[MAXN]; int source,sink; inline void insert(int from,int to,int val) {
    e[eid].v=to;
    e[eid].val=val;
    e[eid].next=p[from];
    p[from]=eid++;
 
    e[eid].v=from;
    e[eid].val=0;
    e[eid].next=p[to];
    p[to]=eid++; } void init() {
    scanf("%d %d", &n, &m); int i, P, S, E, j;
    memset(p, -1, sizeof(p));
    ans = eid = 0; for (i = 1; i <= n; ++i) {
        scanf("%d %d %d", &P, &S, &E);
        ans += P;
        insert(0, i, P); for (j = S; j <= E; ++j) {
            insert(i, n+j, 1); } } for (i = 1; i <= 500; ++i) {
        insert(i+n, n+500+1, m); } } inline int dfs(int pos,int cost, int cnt) { if (pos==sink) { return cost; } int j,minh=cnt-1,lv=cost,d; for (j=p[pos];j!=-1;j=e[j].next) { int v=e[j].v,val=e[j].val; if(val>0) { if (h[v]+1==h[pos]) { if (lv<e[j].val) d=lv; else d=e[j].val;
 
                d=dfs(v,d,cnt);
                e[j].val-=d;
                e[j^1].val+=d;
                lv-=d; if (h[source]>=cnt) return cost-lv; if (lv==0) break; } if (h[v]<minh)    minh=h[v]; } } if (lv==cost) { --gap[h[pos]]; if (gap[h[pos]]==0) h[source]=cnt;
        h[pos]=minh+1; ++gap[h[pos]]; } return cost-lv; } int sap(int st,int ed, int cnt) {
    source=st;
    sink=ed; int ret=0;
    memset(gap,0,sizeof(gap));
    memset(h,0,sizeof(h));
 
    gap[st]=cnt; while (h[st]<cnt) {
        ret+=dfs(st,INT_MAX, cnt); } return ret; } int main() { int T, cas = 0;
    scanf("%d", &T); while (T--) {
        init(); bool flag = false; if (ans == sap(0, n+500+1, n+500+2)) {
            flag = true; }
        printf("Case %d: %s\n\n", ++cas, flag ? "Yes" : "No"); } return 0; 

}