Drainage Ditches
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9715    Accepted Submission(s): 4623


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



 


题意：给定m条边和n个顶点（从1開始）。边为（u。v，c）源点是1，汇点是n。求最大流。




题解：Dinic + 链式前向星，新模板get.






#include <stdio.h>
#include <string.h>

#define maxn 205
#define maxm 410
#define inf 0x3f3f3f3f

int head[maxn], n, m, source, sink, id; // n个点m条边
struct Node {
    int u, v, c, next;
} E[maxm];
int que[maxn], pre[maxn], Layer[maxn];
bool vis[maxn];

void addEdge(int u, int v, int c) {
    E[id].u = u; E[id].v = v;
    E[id].c = c; E[id].next = head[u];
    head[u] = id++;

    E[id].u = v; E[id].v = u;
    E[id].c = 0; E[id].next = head[v];
    head[v] = id++;
}

void getMap() {
    int u, v, c; id = 0;
    memset(head, -1, sizeof(int) * (n + 1));
    source = 1; sink = n;
    while(m--) {
        scanf("%d%d%d", &u, &v, &c);
        addEdge(u, v, c);
    }
}

bool countLayer() {
    memset(Layer, 0, sizeof(int) * (n + 1));
    int id = 0, front = 0, u, v, i;
    Layer[source] = 1; que[id++] = source;
    while(front != id) {
        u = que[front++];
        for(i = head[u]; i != -1; i = E[i].next) {
            v = E[i].v;
            if(E[i].c && !Layer[v]) {
                Layer[v] = Layer[u] + 1;
                if(v == sink) return true;
                else que[id++] = v;
            }
        }
    }
    return false;
}

int Dinic() {
    int i, u, v, minCut, maxFlow = 0, pos, id = 0;
    while(countLayer()) {
        memset(vis, 0, sizeof(bool) * (n + 1));
        memset(pre, -1, sizeof(int) * (n + 1));
        que[id++] = source; vis[source] = 1;
        while(id) {
            u = que[id - 1];
            if(u == sink) {
                minCut = inf;
                for(i = pre[sink]; i != -1; i = pre[E[i].u])
                    if(minCut > E[i].c) {
                        minCut = E[i].c; pos = E[i].u;
                    }
                maxFlow += minCut;
                for(i = pre[sink]; i != -1; i = pre[E[i].u]) {
                    E[i].c -= minCut;
                    E[i^1].c += minCut;
                }
                while(que[id-1] != pos)
                    vis[que[--id]] = 0;
            } else {
                for(i = head[u]; i != -1; i = E[i].next)
                    if(E[i].c && Layer[u] + 1 == Layer[v = E[i].v] && !vis[v]) {
                        vis[v] = 1; que[id++] = v; pre[v] = i; break;
                    }
                if(i == -1) --id;
            }
        }
    }
    return maxFlow;
}

void solve() {
    printf("%d\n", Dinic());
}

int main() {
    while(scanf("%d%d", &m, &n) == 2) {
        getMap();
        solve();
    }
}




    
        
         
            
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