HDU-4966 GGS-DDU （最小树形图(有向图的最小生成树)[朱刘算法]）
GGS-DDU
http://acm.hdu.edu.cn/showproblem.php?pid=4966
Problem Description
Do you think this is a strange problem name? That is because you don't know its full name---'Good Good Study and Day Day Up!". Very famous sentence! Isn't it?
Now "GGS-DDU" is lzqxh's target! He has N courses and every course is divided into a plurality of levels. Just like College English have Level 4 and Level 6.
To simplify the problem, we suppose that the i-th course has Levels from level 0 to level a[i]. And at the beginning, lzqxh is at Level 0 of every course. Because his target is "GGS-DDU", lzqxh wants to reach the highest Level of every course.
Fortunately, there are M tutorial classes. The i-th tutoial class requires that students must reach at least Level L1[i] of course c[i] before class begins. And after finishing the i-th tutorial class, the students will reach Level L2[i] of course d[i]. The i-th tutoial class costs lzqxh money[i].
For example, there is a tutorial class only students who reach at least Level 5 of "Tiyu" can apply. And after finishing this class, the student's "MeiShu" will reach Level 10 if his "MeiShu"'s Level is lower than 10. (Don't ask me why! Supernatural class!!!")
Now you task is to help lzqxh to compute the minimum cost!
Now "GGS-DDU" is lzqxh's target! He has N courses and every course is divided into a plurality of levels. Just like College English have Level 4 and Level 6.
To simplify the problem, we suppose that the i-th course has Levels from level 0 to level a[i]. And at the beginning, lzqxh is at Level 0 of every course. Because his target is "GGS-DDU", lzqxh wants to reach the highest Level of every course.
Fortunately, there are M tutorial classes. The i-th tutoial class requires that students must reach at least Level L1[i] of course c[i] before class begins. And after finishing the i-th tutorial class, the students will reach Level L2[i] of course d[i]. The i-th tutoial class costs lzqxh money[i].
For example, there is a tutorial class only students who reach at least Level 5 of "Tiyu" can apply. And after finishing this class, the student's "MeiShu" will reach Level 10 if his "MeiShu"'s Level is lower than 10. (Don't ask me why! Supernatural class!!!")
Now you task is to help lzqxh to compute the minimum cost!
Input
The input contains multiple test cases.
The first line of each case consists of two integers, N (N<=50) and M (M<=2000).
The following line contains N integers, representing a[1] to a[N]. The sum of a[1] to a[N] will not exceed 500.
The next M lines, each have five integers, indicating c[i], L1[i], d[i], L2[i] and money[i] (1<=c[i], d[i]<=N, 0<=L1[i]<=a[c[i]], 0<=L2[i]<=a[d[i]], money[i]<=1000) for the i-th tutorial class. The courses are numbered from 1 to N.
The input is terminated by N = M = 0.
The first line of each case consists of two integers, N (N<=50) and M (M<=2000).
The following line contains N integers, representing a[1] to a[N]. The sum of a[1] to a[N] will not exceed 500.
The next M lines, each have five integers, indicating c[i], L1[i], d[i], L2[i] and money[i] (1<=c[i], d[i]<=N, 0<=L1[i]<=a[c[i]], 0<=L2[i]<=a[d[i]], money[i]<=1000) for the i-th tutorial class. The courses are numbered from 1 to N.
The input is terminated by N = M = 0.
Output
Output the minimum cost for achieving lzqxh's target in a line. If his target can't be achieved, just output -1.
Sample Input
3 4 3 3 1 1 0 2 3 10 2 1 1 2 10 1 2 3 1 10 3 1 1 3 10 0 0
Sample Output
40
题目大意：有n个课程，每个课程又等级a[i]，初始都在等级0，满足c[j]课程等级大于等于L1[j]时，可花费money[j]使得d[j]课程等级达到L2[j]，求最少需要多少钱？
把每个课程点每一个等级看成一个点，则题目给点条件就可以看成有向边，再添上每一个课程高等级到低等级点权值为0的边，就可以使c[j]课程等级大于L1[j]时，仍可花费money[j]达到d[j]课程等级L2[j]，再添加一个虚拟点根结点，使其能无花费到达每一个课程的0等级结点。
这样就可以直接用 最小树形图 点模板了。
最开始一直WA，最后把MAXN从505换成555就AC了，然后就反应过来最大算错了（不提示越界很是费解） : (
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN=555;
const int MAXM=2555;
const int INF=0x3f3f3f3f;
int n,m;
int pre[MAXN],id[MAXN],vis[MAXN];//pre[i]表示以i点为终点的最小权值入边的起点，id[i]表示i点在新图上对应的点
int inEdge[MAXN];//inEdge[i]表示以i点为终点的入边的最小权值
struct Edge {
int s,e;
int w;
}edge[MAXM];
int Directed_MST(int root) {
int s,e;
int ans=0;
while(true) {
for(int i=0;i<n;++i)
inEdge[i]=INF;
for(int i=0;i<m;++i) {//找权值最小的入边
s=edge[i].s;
e=edge[i].e;
if(edge[i].w<inEdge[e]&&s!=e) {//如果当前边非自环且权值更小
pre[e]=s;
inEdge[e]=edge[i].w;
}
}
for(int i=0;i<n;++i)//如果存在一点没有入边则无法形成最小树形图
if(i!=root&&inEdge[i]==INF)
return -1;
int cnt=0;//cnt表示新图的点数
memset(id,-1,sizeof(id));
memset(vis,-1,sizeof(vis));
inEdge[root]=0;
for(int i=0;i<n;++i) {
ans+=inEdge[i];
int cur=i;
while(vis[cur]!=i&&cur!=root&&id[cur]==-1) {
vis[cur]=i;
cur=pre[cur];
}
if(cur!=root&&id[cur]==-1) {//找到一个环
e=pre[cur];
while(e!=cur) {//在一个环上的点都映射到新图上的同一个点
id[e]=cnt;
e=pre[e];
}
id[cur]=cnt++;
}
}
if(cnt==0)//无环则返回答案
return ans;
for(int i=0;i<n;++i)//每个点映射到新图的一个点
if(id[i]==-1)
id[i]=cnt++;
for(int i=0;i<m;++i) {//建立新图
s=edge[i].s;
e=edge[i].e;
edge[i].s=id[s];
edge[i].e=id[e];
if(id[s]!=id[e])
edge[i].w-=inEdge[e];
}
n=cnt;//更新新图的点数
root=id[root];//更新新图的根
}
return ans;
}
int main() {
int a,c,L1,d,L2,money,num;
int index[55][505];//index为每个课程点每个等级建立一个点
while(scanf("%d%d",&n,&m),n!=0||m!=0) {
num=1;
for(int i=1;i<=n;++i) {//每个课程每个等级映射到一个点
scanf("%d",&a);
for(int j=0;j<=a;++j)
index[i][j]=num++;
index[i][502]=a;
}
for(int i=0;i<m;++i) {
scanf("%d%d%d%d%d",&c,&L1,&d,&L2,&money);
edge[i].s=index[c][L1];
edge[i].e=index[d][L2];
edge[i].w=money;
}
for(int i=1;i<=n;++i) {
for(int j=1;j<=index[i][502];++j) {//每一课程相邻的两个等级，高等级到低等级有一条权值为0点有向边
edge[m].s=index[i][j];
edge[m].e=index[i][j-1];
edge[m++].w=0;
}
}
for(int i=1;i<=n;++i) {//建立一个虚拟树根，到每个课程的0等级都有一条权值为0的有向边
edge[m].s=0;
edge[m].e=index[i][0];
edge[m++].w=0;
}
n=num;
printf("%d\n",Directed_MST(0));
}
return 0;
}