bzoj 1003: [ZJOI2006]物流运输trans

这道题还算比较水。。

spfa预处理从第i天到第j天的可行最短路存为t[i][j]，

然后直接dp。

f[i] 代表到第i天的最小费用

f[i] = min(f[i],min(t[1][i]*i,f[j]+k+t[j+1][i]*(i-j)))

  1 /*
  2 ID:zsy99021
  3 PROB:bzoj1003 
  4 LANG:C++
  5 */
  6 #include <cstdio>
  7 #include <cstring>
  8 #include <algorithm>
  9 #include <cmath>
 10 #include <iostream>
 11 #include <fstream>
 12 #include <ctime>
 13 #define N 108
 14 #define M 28
 15 #define mid(l,r) ((l+r) >> 1)
 16 #define INF 0x7ffffff
 17 using namespace std;
 18   
 19 long long n,t[N][N],m,k,e,head[M];
 20 long long f[N]; 
 21 bool flag[N][M];
 22   
 23 struct wulala
 24 {
 25     int node,next,dist;
 26 }h[M*M];
 27   
 28 int spfa(int a,int b)
 29 {
 30     bool block[M],inq[M];
 31     int dis[M];
 32     int tail = 0,headd = 1,q[M*M];
 33     memset(block,0,sizeof(block));
 34     memset(inq,0,sizeof(inq));
 35     memset(q,0,sizeof(q));
 36     for (int i = 1;i <= m;i++) dis[i] = INF;
 37     for (int i = a;i <= b;i++)
 38         for (int j = 1;j <= m;j++) if (flag[i][j]) block[j] = true;
 39     q[1] = 1; inq[1] = true; dis[1] = 0;
 40     while(tail < headd)
 41     {
 42         int u = q[++tail];
 43         int c = head[u];
 44         while(c)
 45         {
 46             if (block[h[c].node]||dis[h[c].node] < dis[u] + h[c].dist)
 47             {
 48                 c = h[c].next;
 49                 continue;
 50             }
 51             if (!inq[h[c].node]) q[++headd] = h[c].node;
 52             dis[h[c].node] = dis[u] + h[c].dist;
 53             inq[h[c].node] = true;
 54             c = h[c].next; 
 55         }
 56         inq[u] = false;
 57     }
 58     return(dis[m]);
 59 }
 60   
 61 void init()
 62 {
 63     memset(flag,0,sizeof(flag));
 64     scanf("%d%d%d%d",&n,&m,&k,&e);
 65     for (int i = 1;i <= e;i++)
 66     {
 67         int x,y,z;
 68         scanf("%d%d%d",&x,&y,&z);
 69         h[2*i-1].node = y;
 70         h[2*i].node = x;
 71         h[2*i-1].dist = h[2*i].dist = z;
 72         h[2*i-1].next = head[x];
 73         head[x] = 2 * i - 1;
 74         h[2*i].next = head[y];
 75         head[y] = 2 * i;
 76     }
 77     int d;
 78     scanf("%d",&d);
 79     for (int i = 1;i <= d;i++)
 80     {
 81         int a,b,p;
 82         scanf("%d%d%d",&p,&a,&b);
 83         for (int j = a;j <= b;j++) flag[j][p] = true; 
 84     } 
 85     for (int i = 1;i <= n;i++)
 86         for (int j = i;j <= n;j++)
 87             t[i][j] = spfa(i,j);
 88 }
 89   
 90 void work()
 91 {
 92     memset(f,127,sizeof(f)); 
 93     for (int i = 1;i <= n;i++) f[0] = 0;
 94     for (int i = 1;i <= n;i++)
 95         for (int j = 0;j < i;j++)
 96             f[i] = min(f[i],min(t[1][i]*i,f[j]+k+t[j+1][i]*(i-j)));
 97 }
 98   
 99 void debug()
100 {
101     for (int i = 1;i <= n;i++)
102     {
103         for (int j = 1;j <= n;j++) printf("%d ",t[i][j]);
104         printf("\n");
105     }
106 }
107   
108 int main()
109 {
110     init();
111 //  debug();
112     work();
113     printf("%d\n",f[n]);
114     return 0;
115 }
View Code

 

posted @ 2014-01-05 19:09  乌拉拉979  阅读(239)  评论(0)  编辑  收藏  举报