UVALive - 3487 Duopoly(最小割)

题目大意：有两个公司A和B在申请一些资源
现在给出两个公司所申请的内容，内容包括价钱和申请的资源
现在你做为官方，你只能拒绝一个申请或者接受一个申请，同一个资源不能两个公司都拥有，且申请的资源不能只给部分
问：作为官方，你能得到的最大利益是多少

解题思路：这题的矛盾在于，同一个资源不能两家公司共享，既然如此的话，那就找出矛盾的申请，让他们之间连条线就可以了，容量为INF
设立源点，源点和A公司的申请相连接，容量为申请的价钱
设立汇点，汇点和B公司的申请相连接，容量为申请的价钱
然后找出矛盾的申请，连线。因为只有矛盾的申请才能连线，且只有连线的才能把流流到汇点，所以跑出来的最大流就是我们所需要的最小割
所以答案就是所有申请的价钱总和 - 最小割
如果不懂的话，请百度最小割，去掉最小割后，两个点集之间就没有关系，也就不会存在矛盾了

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;#define M 1000010#define N 10010#define INF 0x3f3f3f3fstruct Edge{    int u, v, cap, flow, next;}E[M];struct Dinic{    int head[N], d[N];    int tot, sink, source;    void init() {        memset(head, -1, sizeof(head));        tot = 0;    }    inline void AddEdge(int u, int v, int cap) {        E[tot].u = u; E[tot].v = v; E[tot].cap = cap; E[tot].flow = 0; E[tot].next = head[u]; head[u] = tot++;        u = u ^ v; v = u ^ v; u = u ^ v;        E[tot].u = u; E[tot].v = v; E[tot].cap = 0; E[tot].flow = 0; E[tot].next = head[u]; head[u] = tot++;    }    inline bool bfs(int s) {        int u, v;        memset(d, 0, sizeof(d));        queue<int> Q;        Q.push(s);        d[s] = 1;        while (!Q.empty()) {            u = Q.front(); Q.pop();            if (u == sink) return true;            for (int i = head[u]; ~i; i = E[i].next) {                v = E[i].v;                if (!d[v] && E[i].cap - E[i].flow > 0) {                    d[v] = d[u] + 1;                    Q.push(v);                }            }        }        return false;    }    int dfs(int x, int a) {        if (x == sink || a == 0)            return a;        int f, flow = 0;        for (int i = head[x]; ~i; i = E[i].next) {            int v = E[i].v;            if (d[v] == d[x] + 1 && E[i].cap - E[i].flow > 0) {                f = dfs(v, min(a, E[i].cap - E[i].flow));                E[i].flow += f;                E[i^1].flow -= f;                flow += f;                a -= f;                if (!a) break;            }        }        if (flow == 0) d[x] = 0;        return flow;    }    int Maxflow(int source, int sink) {        int flow = 0;        this->sink = sink;        while (bfs(source)) flow += dfs(source, INF);        return flow;    }};Dinic dinic;#define maxn 3010#define S 300010int PriceTel[maxn], PriceMob[maxn];int ChanTel[S], ChanMob[S];bool vis[maxn][maxn];int n, m, cas = 1;void init() {    memset(ChanTel, 0, sizeof(ChanTel));    memset(ChanMob, 0, sizeof(ChanMob));    memset(vis, 0, sizeof(vis));    int Max = -INF, ans = 0;    char c;    scanf("%d", &n);    for (int i = 1; i <= n; i++) {        scanf("%d", &PriceTel[i]);        ans += PriceTel[i];        c = getchar();        while (c != '\n') {            int t = 0;            while (c = getchar(), c >= '0' && c <= '9')                 t = t * 10 + c - '0';            Max = max(Max, t);            ChanTel[t] = i;        }    }    scanf("%d", &m);    for (int i = 1; i <= m; i++) {        scanf("%d", &PriceMob[i]);        ans += PriceMob[i];        c = getchar();        while (c != '\n') {            int t = 0;            while (c = getchar(),  c >= '0' && c <= '9')                t = t * 10 + c - '0';            Max = max(Max, t);            ChanMob[t] = i;        }    }    int source = 0, sink = n + m + 1;    dinic.init();    for (int i = 1; i <= n; i++)        dinic.AddEdge(source, i, PriceTel[i]);    for (int i = 1; i <= m; i++)        dinic.AddEdge(i + n, sink, PriceMob[i]);    for (int i = 1; i <= Max; i++) {        if (!ChanTel[i] || !ChanMob[i] || vis[ChanTel[i]][ChanMob[i]]) continue;        vis[ChanTel[i]][ChanMob[i]] = true;        dinic.AddEdge(ChanTel[i], ChanMob[i] + n, INF);    }    int maxflow = dinic.Maxflow(source, sink);    printf("Case %d:\n", cas++);    printf("%d\n", ans - maxflow);}int main() {    int test;    scanf("%d", &test);    bool flag = false;    while (test--) {        if (flag)            printf("\n");        flag = true;        init();    }    return 0;}