BZOJ1520: [POI2006]Szk-Schools

1520: [POI2006]Szk-Schools

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 342  Solved: 173
[Submit][Status]

Description

Input

Output

如果有可行解, 输出最小代价,否则输出NIE.

Sample Input

5
1 1 2 3
1 1 5 1
3 2 5 5
4 1 5 10
3 3 3 1

Sample Output

9

HINT

Source

题解：

裸的最小费用最大流。。。

代码：

  1 #include<cstdio>
  2 
  3 #include<cstdlib>
  4 
  5 #include<cmath>
  6 
  7 #include<cstring>
  8 
  9 #include<algorithm>
 10 
 11 #include<iostream>
 12 
 13 #include<vector>
 14 
 15 #include<map>
 16 
 17 #include<set>
 18 
 19 #include<queue>
 20 
 21 #include<string>
 22 
 23 #define inf 1000000000
 24 
 25 #define maxn 500+100
 26 
 27 #define maxm 250000
 28 
 29 #define eps 1e-10
 30 
 31 #define ll long long
 32 
 33 #define pa pair<int,int>
 34 
 35 #define for0(i,n) for(int i=0;i<=(n);i++)
 36 
 37 #define for1(i,n) for(int i=1;i<=(n);i++)
 38 
 39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 40 
 41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 42 
 43 #define mod 1000000007
 44 
 45 using namespace std;
 46 
 47 inline int read()
 48 
 49 {
 50 
 51     int x=0,f=1;char ch=getchar();
 52 
 53     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
 54 
 55     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
 56 
 57     return x*f;
 58 
 59 }
 60 int n,m,k,mincost,maxflow,tot=1,s,t,head[maxn],q[maxm],d[maxn],from[2*maxm];
 61 
 62 bool v[maxn];
 63 
 64 struct edge{int from,next,go,v,c;}e[2*maxm];
 65 
 66 void ins(int x,int y,int z,int w)
 67 
 68 {
 69 
 70     e[++tot].go=y;e[tot].from=x;e[tot].v=z;e[tot].c=w;e[tot].next=head[x];head[x]=tot;
 71 
 72 }
 73 
 74 void insert(int x,int y,int z,int w)
 75 
 76 {
 77 
 78     ins(x,y,z,w);ins(y,x,0,-w);
 79 
 80 }
 81 
 82 bool spfa()
 83 
 84 {
 85 
 86     for (int i=s;i<=t;i++){v[i]=0;d[i]=inf;}
 87 
 88     int l=0,r=1,y;q[1]=s;d[s]=0;v[0]=1;
 89 
 90     while(l!=r)
 91 
 92     {
 93 
 94         int x=q[++l];if(l==maxn)l=0;v[x]=0;
 95 
 96         for (int i=head[x];i;i=e[i].next)
 97 
 98          if(e[i].v&&d[x]+e[i].c<d[y=e[i].go])
 99 
100          {
101 
102             d[y]=d[x]+e[i].c;from[y]=i;
103 
104             if(!v[y]){v[y]=1;q[++r]=y;if(r==maxn)r=0;}
105 
106          }
107 
108     }
109 
110     return d[t]!=inf;
111 
112 }
113 
114 void mcf()
115 
116 {
117 
118     while(spfa())
119 
120     {
121 
122         int tmp=inf;
123 
124         for(int i=from[t];i;i=from[e[i].from]) tmp=min(tmp,e[i].v);
125 
126         mincost+=d[t]*tmp;maxflow+=tmp;
127 
128         for(int i=from[t];i;i=from[e[i].from]){e[i].v-=tmp;e[i^1].v+=tmp;}
129 
130     }
131 
132 }
133 
134 int main()
135 
136 {
137 
138     freopen("input2.txt","r",stdin);
139 
140     freopen("output3.txt","w",stdout);
141 
142     n=read();s=0;t=2*n+1;
143     for1(i,n)
144     {
145         insert(s,i,1,0);insert(i+n,t,1,0);
146         int x=read(),l=read(),r=read(),y=read();
147         for2(j,l,r)insert(i,j+n,1,y*(abs(j-x)));
148     }
149     mcf();
150     if(maxflow==n)printf("%d\n",mincost);else puts("NIE");
151 
152     return 0;
153 
154 }
View Code

 

posted @ 2014-09-17 17:58  ZYF-ZYF  Views(227)  Comments(0)  Edit  收藏  举报