Drainage Ditches
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 37500 |
|
Accepted: 13862 |
Description
Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.
Input
The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.
Output
For each case, output a single integer, the maximum rate at which water may emptied from the pond.
Sample Input
5 4
1 2 40
1 4 20
2 4 20
2 3 30
3 4 10
Sample Output
50
話說這是道網絡流的模版題，鄙人沒寫過多少網絡流，于是忍不住來練練手
邪惡的是，我打了模版之后，才發現我模版好多錯誤（自己改的模版），悲劇
然后終于改好了，然后一直wa，
不解，重邊我也處理了呀，怎么搞的，
然后去看discuss，忽然發現一組數據
input：
3 2
1 2 3
1 2 4
1 2 5
output：
12
頓時無語了，原來重邊在這里要相加啊，我開始取得min，后來取得max
呃，表示自己真心是水貨
1
#include<stdio.h>
2#include<string.h>
3#include<math.h>
4#define MAX 205
5#define inf 0x7fffffff
6int n,m,s,t;
7int map[MAX][MAX];
8int d[MAX],r[MAX][MAX],num[MAX],pre[MAX];
9int min(int a,int b)
10
{
11
if(a<b) return a;
12 else return b;
13
}
14int void bfs()
15{
16 int i,k;
17 int q[MAX*MAX],head,tail;
18 for(i=1;i<=n;i++)
19 d[i]=n+1;
20 memset(num,0,sizeof(num));
21 head=0;
22 tail=1;
23 q[tail]=t;
24 d[t]=0;
25 num[0]=1;
26 while(head<tail)
27
{
28 head++;
29 k=q[head];
30 for(i=1; i<=n; i++)
31 {
32 if(d[i]>n&&r[i][k]>0)
33 {
34 d[i]=d[k]+1;
35 tail++;
36 q[tail]=i;
37 num[d[i]]++;
38
}
39 }
40 }
41}
42int findalowarc(int i)
43{
44 int j;
45 for(j=1; j<=n; j++)
46 if((r[i][j]>0)&&(d[i]==d[j]+1)) return j;
47 return -1;
48}
49int relabel(int i)
50{
51 int j;
52 int mm=inf;
53 for(j=1; j<=n; j++)
54 {
55 if (r[i][j]>0) mm=min(mm,d[j]+1);
56 }
57 return (mm==inf)?n:mm;
58}
59int maxflow()
60{
61 int flow,i,j,delta;
62 int x;
63 bfs();
64 i=s;
65 flow=0;
66 memset(pre,-1,sizeof(pre));
67 while(d[s]<=n)
68 {
69 j=findalowarc(i);
70 if(j>0)
71 {
72 pre[j]=i;
73 i=j;
74 if(i==t)
75 {
76 delta=inf;
77 for(i=t; i!=s; i=pre[i])
78 delta=min(delta,r[pre[i]][i]);
79 for(i=t; i!=s; i=pre[i])
80 {
81 r[pre[i]][i]-=delta;
82 r[i][pre[i]]+=delta;
83 }
84 flow+=delta;
85 }
86 }
87 else
88 {
89 x=relabel(i);
90 num[x]++;
91 num[d[i]]--;
92 if(num[d[i]]==0) return flow;//間隙優化
93 d[i]=x;
94 if(i!=s) i=pre[i];
95 }
96 }
97 return flow;
98}
99int main()
100{
101 int i,j;
102 int x,y,z;
103 while (scanf("%d%d",&m,&n)!=EOF)
104 {
105 s=1;
106 t=n;
107 memset(map,0,sizeof(map));
108 for (i=1; i<=m; i++ )
109 {
110 scanf("%d%d%d",&x,&y,&z);
111 if (map[x][y]==0)
112 {
113 map[x][y]=z;
114 }
115 else
116 {
117 map[x][y]=map[x][y]+z;
118 }
119 }
120 for(i=1; i<=n; i++)
121 for(j=1; j<=n; j++)
122 r[i][j]=map[i][j];
123 printf("%d\n",maxflow());
124 }
125 return 0;
126}
127
還發現個問題
我寫的sap怎么跑了16ms，同學寫的ek才跑了0ms
why？？