[2014Contest_2I]The Worst Schedule

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解题思路：

涨姿势，什么叫闭合图： http://www.cnblogs.com/wuyiqi/archive/2012/03/12/2391960.html

第一种解法：

第一问：s和i点连一个权为ai的边，y和i点连一个权为bi的边，若j依赖i，则i向j连一条权为inf的边。对图求最小割。

图可以这样建的原因是：假设j依赖i，若i到y的边断了，即i为延迟完成。这时候从s到i的的残留会转移到i->j->y，这时候也只能把j到y的边断了。

第二问：s集合是延迟集合，t是提前集合。把s的点都跑一遍标记一下，剩余的都可以分给t集。

第二种解法：

第一问：我也不知道他们是怎么想到的，先对所有点取sum(min(del[i], adv[i]))并把它当作答案。如果没有依赖的话，显然这个就是答案。

对于所有点，如果adv[i] - del[i] > 0（表示提前完成比延迟完成多的代价），则s到i连一条adv[i] - del[i]边（表示延迟完成比提前完成多的代价），否则i到t连一条边del[i] - adv[i]，若j依赖i，则i向j连一条INF的边。对上图求最小割。

假设存在j依赖i，并且del[i] - adv[i] < 0，之前我们默认i是延迟的，若把s->i断掉，则i变成是提前的；若不把s->i断掉，则残留压向i->j，若存在j->t的边，则j->t也要断掉。符合题目要求。

太机智了。

第二问：同上。

第一种：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <map>#include <set>#include <vector>#include <utility>#include <queue>#include <stack>#include <cstdlib>#include <ctime>using namespace std;#pragma comment(linker,"/STACK:102400000,102400000")#define LL long long#define ULL unsigned long long#define Hei cout << "Czy!!!" << endl;#define lson rt << 1, l, mid#define rson  rt << 1 | 1, mid + 1, r#define MOD 1000000007/*clock_t t1, t2;t1 = clock();t2 = clock();cout << (double)(t2 - t1) / CLOCKS_PER_SEC << endl;*/#define maxn 210#define INF 100000000struct Edge{    int from, to, cap, flow;    Edge(int from = 0, int to = 0, int cap = 0, int flow = 0): from(from), to(to), cap(cap), flow(flow) {}};vector<int> G[maxn];vector<Edge> edges; bool vis[210];int tot;void AddEdge(int from, int to, int cap){    edges.push_back(Edge(from, to, cap, 0));    edges.push_back(Edge(to, from, 0, 0));    int m = edges.size();    G[from].push_back(m - 2);    G[to].push_back(m - 1);}struct Dinic{    int s, t;    bool vis[maxn];    int d[maxn];    int cur[maxn];    bool BFS()    {        memset(vis, 0, sizeof(vis));        queue<int> Q;        while (!Q.empty()) Q.pop();        Q.push(s);        d[s] = 0;        vis[s] = 1;        while (!Q.empty())        {            int x = Q.front();            Q.pop();            for (int i = 0; i < G[x].size(); i++)            {                Edge &e = edges[G[x][i]];                if (!vis[e.to] && e.cap > e.flow)                {                    vis[e.to] = 1;                    d[e.to] = d[x] + 1;                    Q.push(e.to);                }            }        }        return vis[t];    }    int DFS(int x, int a)    {        if (x == t || a == 0) return a;        int flow = 0, f ;        for (int& i = cur[x]; i < G[x].size(); i++)        {            Edge& e = edges[G[x][i]];            if (d[x] + 1 == d[e.to] &&  (f = DFS(e.to, min(a, e.cap - e.flow))) > 0)            {                e.flow += f;                edges[G[x][i] ^ 1].flow -= f;                flow += f;                a -= f;                if (a == 0) break;            }        }        return flow;    }    int MaxFlow(int s, int t)    {        this->s = s;        this->t = t;        int flow = 0;        while (BFS())        {            //Hei;            memset(cur, 0, sizeof(cur));            flow += DFS(s, INF);        }        return flow;    }}solver;void dfs(int u){    vis[u] = true;    for (int i = 0; i < G[u].size(); i++)    {        Edge e = edges[G[u][i]];        if (e.flow != e.cap && !vis[e.to]) dfs(e.to);    }}int main(){    freopen("gen.in", "r", stdin);    freopen("my.out", "w", stdout);    int n;    while (cin >> n)    {        edges.clear();        for (int i = 0; i <= n + 1; i++) G[i].clear();        for (int i = 1; i <= n; i++)        {            int x;            scanf("%d", &x);            AddEdge(0, i, x);        }        for (int i = 1; i <= n; i++)        {            int x;            scanf("%d", &x);            AddEdge(i, n + 1, x);        }        int m;        cin >> m;        while (m--)        {            int x, y;            scanf("%d%d", &x, &y);            AddEdge(x, y, INF);        }        cout << solver.MaxFlow(0, n + 1) << " ";        memset(vis, false, sizeof(vis));        dfs(0);        int tot = 0;        for (int i = 1; i <= n; i++) if (!vis[i]) tot++;        cout << tot << endl;    }    fclose(stdin);    fclose(stdout);}

第二种：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <map>#include <set>#include <vector>#include <utility>#include <queue>#include <stack>#include <cstdlib>#include <ctime>using namespace std;#pragma comment(linker,"/STACK:102400000,102400000")#define LL long long#define ULL unsigned long long#define Hei cout << "Czy!!!" << endl;#define lson rt << 1, l, mid#define rson  rt << 1 | 1, mid + 1, r#define MOD 1000000007/*clock_t t1, t2;t1 = clock();t2 = clock();cout << (double)(t2 - t1) / CLOCKS_PER_SEC << endl;*/#define maxn 210#define INF 100000000struct Edge{    int from, to, cap, flow;    Edge(int from = 0, int to = 0, int cap = 0, int flow = 0): from(from), to(to), cap(cap), flow(flow) {}};vector<int> G[maxn];vector<Edge> edges;bool vis[210];int adv[210], del[210];void AddEdge(int from, int to, int cap){    edges.push_back(Edge(from, to, cap, 0));    edges.push_back(Edge(to, from, 0, 0));    int m = edges.size();    G[from].push_back(m - 2);    G[to].push_back(m - 1);}struct Dinic{    int s, t;    bool vis[maxn];    int d[maxn];    int cur[maxn];    bool BFS()    {        memset(vis, 0, sizeof(vis));        queue<int> Q;        while (!Q.empty()) Q.pop();        Q.push(s);        d[s] = 0;        vis[s] = 1;        while (!Q.empty())        {            int x = Q.front();            Q.pop();            for (int i = 0; i < G[x].size(); i++)            {                Edge &e = edges[G[x][i]];                if (!vis[e.to] && e.cap > e.flow)                {                    vis[e.to] = 1;                    d[e.to] = d[x] + 1;                    Q.push(e.to);                }            }        }        return vis[t];    }    int DFS(int x, int a)    {        if (x == t || a == 0) return a;        int flow = 0, f ;        for (int& i = cur[x]; i < G[x].size(); i++)        {            Edge& e = edges[G[x][i]];            if (d[x] + 1 == d[e.to] &&  (f = DFS(e.to, min(a, e.cap - e.flow))) > 0)            {                e.flow += f;                edges[G[x][i] ^ 1].flow -= f;                flow += f;                a -= f;                if (a == 0) break;            }        }        return flow;    }    int MaxFlow(int s, int t)    {        this->s = s;        this->t = t;        int flow = 0;        while (BFS())        {            //Hei;            memset(cur, 0, sizeof(cur));            flow += DFS(s, INF);        }        return flow;    }}solver;void dfs(int u){    vis[u] = true;    for (int i = 0; i < G[u].size(); i++)    {        Edge e = edges[G[u][i]];        if (e.flow != e.cap && !vis[e.to]) dfs(e.to);    }}int main(){    freopen("gen.in", "r", stdin);    freopen("my.out", "w", stdout);    int n;    while (cin >> n)    {        edges.clear();        for (int i = 0; i <= n + 1; i++) G[i].clear();        for (int i = 1; i <= n; i++) scanf("%d", &adv[i]);        int ans = 0;        for (int i = 1; i <= n; i++)        {            scanf("%d", &del[i]);            if (adv[i] - del[i] > 0) AddEdge(0, i, adv[i] - del[i]);            else AddEdge(i, n + 1, del[i] - adv[i]);            ans += min(adv[i], del[i]);        }        int m;        cin >> m;        while (m--)        {            int x, y;            scanf("%d%d", &x, &y);            AddEdge(x, y, INF);        }        cout << solver.MaxFlow(0, n + 1) + ans << " ";        memset(vis, false, sizeof(vis));        dfs(0);        int tot = 0;        for (int i = 1; i <= n; i++) if (!vis[i]) tot++;        cout << tot << endl;    }    fclose(stdin);    fclose(stdout);}

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