HDU 6214最少的最小割边数

Posted 2020-11-01 tennant

题目大意如题。

这道题想了很久也没明白题解的做法：

建边的时候每条边权 w = w * (E + 1) + 1;
这样得到最大流 maxflow / (E + 1) ，最少割边数 maxflow % (E + 1)

总之先把它当成黑科技背着吧

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
struct node {
    int from;
    int to;
    int w;
    int next;
}e[160000];
int cur[1500];
int head[1500];
int divv[1500];
int cont, ss, tt;
void add(int from, int to, int w) {
    e[cont].from = from;
    e[cont].w = w;
    e[cont].to = to;
    e[cont].next = head[from];
    head[from] = cont++;
}
int n, m; int makedivv() {
    memset(divv, 0, sizeof(divv));
    divv[ss] = 1;
    queue<int >s;
    s.push(ss);
    while (!s.empty()) {
        int u = s.front();
        if (u == tt)return 1;
        s.pop();
        for (int i = head[u]; i != -1; i = e[i].next) {
            int w = e[i].w;
            int v = e[i].to;
            if (divv[v] == 0 && w) {
                divv[v] = divv[u] + 1;
                s.push(v);
            }
        }
    }
    return 0;
}
int Dfs(int u, int maxflow, int tt) {
    if (u == tt)return maxflow;
    int ret = 0;
    for (int &i = cur[u]; i != -1; i = e[i].next) {
        int v = e[i].to;
        int w = e[i].w;
        if (divv[v] == divv[u] + 1 && w) {
            int f = Dfs(v, min(maxflow - ret, w), tt);
            e[i].w -= f;
            e[i ^ 1].w += f;
            ret += f;
            if (ret == maxflow)return ret;
        }
    }
    return ret;

}
void Dinic() {
    long long int ans = 0;
    while (makedivv() == 1) {
        memcpy(cur, head, sizeof(head));
        ans += Dfs(ss, 0x3f3f3f3f, tt);
    }
    printf("%lld\n", ans % 100000);
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d", &n, &m);
        scanf("%d%d", &ss, &tt);
        cont = 0;
        memset(head, -1, sizeof(head));
        for (int i = 0; i < m; i++) {
            int x, y, z;
            scanf("%d%d%d", &x, &y, &z);
            z = z * 100000 + 1;
            add(x, y, z); add(y, x, 0);
        }
        Dinic();
    }
}