Drainage Ditches
Hal Burch
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. Note however, that there can be more than one ditch between two intersections.
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.
PROGRAM NAME: ditch
INPUT FORMAT
Line 1:
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.
Line 2..N+1:
Each of 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.
SAMPLE INPUT (file ditch.in)
5 4
1 2 40
1 4 20
2 4 20
2 3 30
3 4 10
OUTPUT FORMAT
One line with a single integer, the maximum rate at which water may emptied from the pond.
SAMPLE OUTPUT (file ditch.out)
50
最基本的網(wǎng)絡(luò)流
1: #include<iostream>
2: #include<fstream>
3: #include<string>
4: #include<vector>
5: #include<map>
6: #include<algorithm>
7: #include<sstream>
8: #include <cstring>
9: #include <queue>
10: using namespace std;
11: const int MAXN = 220;
12: const int infi = 0x7FFFFFFF;
13: int capacity[MAXN][MAXN], prenode[MAXN], flow[MAXN];
14: queue<int> mq;
15:
16: int start, end, N;
17: void init(){ 18: freopen("ditch.in","r",stdin); 19: //freopen("e:\\usaco\\ditch.in","r",stdin); 20: start = 1;
21: scanf("%d %d",&N,&end); int c, s, t; 22: memset(capacity,0,sizeof(capacity));
23: for(int i=0;i<N;i++)
24: { 25: scanf("%d %d %d",&c,&s,&t); 26: capacity[c][s] += t; //兩個節(jié)點間不只有一條路
27: }
28: }
29: int bfs(){//尋找增廣路徑 30: while(!mq.empty()) mq.pop();
31: mq.push(start); //源節(jié)點入隊
32: //memset(flow,0,sizeof(flow));
33: memset(prenode,-1,sizeof(prenode)); //重置前向節(jié)點
34: prenode[start] = 0; flow[start]=infi; //源節(jié)點流量無限大
35: while(!mq.empty()){ 36: int cur = mq.front();
37: mq.pop();
38: if(cur == end) break; //到達匯點結(jié)束路徑
39: for(int i=1;i<=end;i++){ 40: if(prenode[i]==-1 && capacity[cur][i]) //訪問當(dāng)前節(jié)點所有未訪問的相鄰節(jié)點，更新flow
41: { 42: prenode[i] = cur;
43: flow[i] = (flow[cur]<capacity[cur][i]?flow[cur]:capacity[cur][i]);
44: mq.push(i);
45: }
46: }
47: }
48: if(prenode[end]==-1) //如果未找到增廣路徑返回-1
49: return -1;
50: return flow[end];
51: }
52: int Edmonds_Karp(){ 53: int total = 0, pathcapacity;//pathcapacity 路徑增加量
54: while((pathcapacity = bfs()) != -1){//可以找到增廣路徑時候進行循環(huán) 55: int cur = end; //從匯點開始增加逆向節(jié)點
56: while( cur != start ){ 57: int t = prenode[cur] ;
58: capacity[t][cur] -= pathcapacity;
59: capacity[cur][t] += pathcapacity;
60: cur = t;
61: }
62: total += pathcapacity;//max_flow
63: }
64: return total;
65: }
66: void output(){ 67: freopen("ditch.out","w",stdout); 68: //freopen("c:\\usaco\\ditch.out","w",stdout); 69: cout<<Edmonds_Karp()<<endl;
70: }
71: int main()
72: { 73: init();
74: output();
75: return 0;
76: }
標(biāo)程：使用貪心法，尋找一條增廣路徑的時候不斷尋找cap最大的，未被訪問的節(jié)點mloc；然后更新跟mloc相鄰的節(jié)點flow以
及prenode信息.最后當(dāng)運行到end時候，更新路徑節(jié)點capacity，同時增加max_flow.重復(fù)上述過程直到找不到增廣路徑
1: #include <stdio.h>
2: #include <string.h>
3:
4: #define MAXI 200
5:
6: /* total drain amount between intersection points */
7: int drain[MAXI][MAXI];
8: int nint; /* number of intersection points */
9:
10: int cap[MAXI]; /* amount of flow that can get to each node */
11: int vis[MAXI]; /* has this node been visited by Dijkstra's yet? */
12: int src[MAXI]; /* the previous node on the path from the source to here */
13:
14: int augment(void)
15: { /* run a Dijkstra's varient to find maximum augmenting path */ 16: int lv;
17: int mloc, max;
18: int t;
19:
20: memset(cap, 0, sizeof(cap));
21: memset(vis, 0, sizeof(vis));
22:
23: cap[0] = 2000000000;
24: while (1)
25: { 26: /* find maximum unvisited node */
27: max = 0;
28: mloc = -1;
29: for (lv = 0; lv < nint; lv++)
30: if (!vis[lv] && cap[lv] > max)
31: { 32: max = cap[lv];
33: mloc = lv;
34: }
35: if (mloc == -1) return 0;
36: if (mloc == nint-1) break; /* max is the goal, we're done */
37:
38: vis[mloc] = -1; /* mark as visited */
39:
40: /* update neighbors, if going through this node improves the
41: capacity */
42: for (lv = 0; lv < nint; lv++)
43: if (drain[mloc][lv] > cap[lv] && max > cap[lv])
44: { 45: cap[lv] = drain[mloc][lv];
46: if (cap[lv] > max) cap[lv] = max;
47: src[lv] = mloc;
48: }
49: }
50: max = cap[nint-1];
51:
52: /* augment path, starting at end */
53: for (lv = nint-1; lv > 0; lv = src[lv])
54: { 55: t = src[lv];
56: drain[t][lv] -= max;
57: drain[lv][t] += max;
58: }
59: return max;
60: }
61:
62: int main(int argc, char **argv)
63: { 64: FILE *fout, *fin;
65: int lv;
66: int num;
67: int p1, p2, c;
68:
69: if ((fin = fopen("ditch.in", "r")) == NULL) 70: { 71: perror ("fopen fin"); 72: exit(1);
73: }
74: if ((fout = fopen("ditch.out", "w")) == NULL) 75: { 76: perror ("fopen fout"); 77: exit(1);
78: }
79:
80: fscanf (fin, "%d %d", &num, &nint);
81: while (num--)
82: { 83: fscanf (fin, "%d %d %d", &p1, &p2, &c);
84: p1--;
85: p2--;
86: drain[p1][p2] += c; /* note += handles two ditches between same points */
87: }
88:
89: /* max flow algorithm: augment while you can */
90: c = 0;
91: while ((p1 = augment()))
92: c += p1;
93: fprintf (fout, "%d\n", c);
94: return 0;
95: }