Drainage Ditches

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem 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

Source

USACO 93

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#define LL long long
#define INF 0x7fffffff
using namespace std;

const int maxn = 205;
int n, m;
int cap[maxn][maxn];
int flow[maxn];
int pre[maxn];

int bfs(int s, int t) {
    memset(pre, -1, sizeof(pre));
    pre[s] = 0;
    flow[s] = INF;

    queue<int> que;
    que.push(s);
    while(!que.empty()) {
        int cur = que.front();
        que.pop();
        if(cur == t) break;		//找到增广路径
        for(int i = 1; i <= m; i++) {
            if(i != s && cap[cur][i] > 0 && pre[i] == -1) {
                pre[i] = cur;	//记录前驱
                flow[i] = min(cap[cur][i], flow[cur]);
                que.push(i);
            }
        }
    }
    if(pre[t] == -1) return -1;
    else return flow[t];
}

int ek(int s, int t) {
    int inc = 0;
    int ans = 0;
    while((inc = bfs(s, t)) != -1) {
        int k = t;
        while(k != s) {
            cap[pre[k]][k] -= inc;
            cap[k][pre[k]] += inc;
            k = pre[k];
        }
        ans += inc;
    }
    return ans;
}

int main() {
    while(cin >> n >> m) {
        memset(cap, 0, sizeof(cap));
        for(int i = 0; i < n; i++) {
            int u, v, w;
            scanf("%d %d %d", &u, &v, &w);
            if(u == v) continue;
            cap[u][v] += w;
        }
        cout << ek(1, m) << endl;
    }
    return 0;
}

Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Problem Description

Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.

Input

The first line of input contains an integer T, denoting the number of test cases.

For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)

Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)

Output

For each test cases, you should output the maximum flow from source 1 to sink N.

Sample Input

2
3 2
1 2 1
2 3 1
3 3
1 2 1
2 3 1
1 3 1

Sample Output

Case 1: 1
Case 2: 2

Author

HyperHexagon

Source

HyperHexagon‘s Summer Gift (Original tasks)

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#define LL long long
#define INF 0x7fffffff
using namespace std;

const int maxn = 205;
int n, m;
int cap[maxn][maxn];
int flow[maxn];
int pre[maxn];

int bfs(int s, int t) {
	memset(pre, -1, sizeof(pre));
	pre[s] = 0;
	flow[s] = INF;

	queue<int> que;
	que.push(s);
	while(!que.empty()) {
		int cur = que.front();
		que.pop();
		if(cur == t) break; //找到增广路径
		for(int i = 1; i <= m; i++) {
			if(i != s && cap[cur][i] > 0 && pre[i] == -1) {
				pre[i] = cur;	//记录前驱
				flow[i] = min(cap[cur][i], flow[cur]);
				que.push(i);
			}
		}
	}
	if(pre[t] == -1) return -1;
	else return flow[t];
}

int ek(int s, int t) {
	int inc = 0;
	int ans = 0;
	while((inc = bfs(s, t)) != -1) {
		int k = t;
		while(k != s) {
			cap[pre[k]][k] -= inc;
			cap[k][pre[k]] += inc;
			k = pre[k];
		}
		ans += inc;
	}
	return ans;
}

int main() {
	int T;
	scanf("%d", &T);
	for(int cas = 1; cas <= T; cas++) {
		cin >> m >> n;
		memset(cap, 0, sizeof(cap));
		for(int i = 0; i < n; i++) {
			int u, v, w;
			scanf("%d %d %d", &u, &v, &w);
			if(u == v) continue;
			cap[u][v] += w;
		}
		cout << "Case " << cas << ": " << ek(1, m) << endl;
	}
	return 0;
}