SCU2016-01 G题（最大流 + floyd_warshall）

分析：

其实就是经典的限制终点的容量的最大流。建立超级源点和汇点。最小化最大路径，二分一下。对于任意最短路Floyd_warshall一下。
code：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>using namespace std;#define pr(x) cout << #x << ": " << x << "  " #define pl(x) cout << #x << ": " << x << endl;const int INF = int(1e9);////////////////////////最大流开始//////////////////////////////////////typedef int cap_type;#define MAX_V 200 + 30 + 16// 用于表示边的结构体（终点、容量、反向边）struct edge{    int to, rev;    cap_type cap;    edge(int to, cap_type cap, int rev) : to(to), cap(cap), rev(rev)    {}};vector <edge> G[MAX_V];   // 图的邻接表表示int level[MAX_V];      // 顶点到源点的距离标号int iter[MAX_V];       // 当前弧，在其之前的边已经没有用了// 向图中加入一条从from到to的容量为cap的边void add_edge(int from, int to, int cap){    G[from].push_back(edge(to, cap, G[to].size()));    G[to].push_back(edge(from, 0, G[from].size() - 1));}// 通过BFS计算从源点出发的距离标号void bfs(int s){    memset(level, -1, sizeof(level));    queue<int> que;    level[s] = 0;    que.push(s);    while (!que.empty())    {        int v = que.front();        que.pop();        for (int i = 0; i < (int)G[v].size(); ++i)        {            edge &e = G[v][i];            if (e.cap > 0 && level[e.to] < 0)            {                level[e.to] = level[v] + 1;                que.push(e.to);            }        }    }}// 通过DFS寻找增广路cap_type dfs(int v, int t, cap_type f){    if (v == t)    {        return f;    }    for (int &i = iter[v]; i < (int)G[v].size(); ++i)    {        edge &e = G[v][i];        if (e.cap > 0 && level[v] < level[e.to])        {            cap_type d = dfs(e.to, t, min(f, e.cap));            if (d > 0)            {                e.cap -= d;                G[e.to][e.rev].cap += d;                return d;            }        }    }    return 0;}// 求解从s到t的最大流cap_type max_flow(int s, int t){    cap_type flow = 0;    for (;;)    {        bfs(s);        if (level[t] < 0)        {            return flow;        }        memset(iter, 0, sizeof(iter));        cap_type f;        while ((f = dfs(s, t, 0x3f3f3f3f) > 0))        {            flow += f;        }    }}///////////////////////////////最大流结束/////////////////////////////////////struct jibancanyang{    int K, C, M, Ga[234][234];    void floyd_warshall() {        for (int i = 0; i < K + C; ++i) {            for (int j = 0; j < K + C; ++j) {                for (int k = 0; k < K + C; ++k) {                    Ga[j][k] = min(Ga[j][i] + Ga[i][k], Ga[j][k]);                }            }        }    }    bool judge(int mid) {        int s = C + K, t = s + 1;        for (int i = 0; i <= t; ++i) G[i].clear();        for (int i = 0; i < K; ++i) add_edge(i, t, M);        for (int i = K; i < K + C; ++i) add_edge(s, i, 1);        for (int i = 0; i < K; ++i) {            for (int j = K; j < K + C; ++j) {                if (Ga[i][j] <= mid) add_edge(j, i, 1);            }        }        return max_flow(s, t) == C;    }    int binary_search() {        floyd_warshall();        int l = 0, r = 200 * (K + C);        while (l < r) {            int mid = (r - l) / 2 + l;            if (judge(mid)) r = mid;            else l = mid + 1;        }        return l;    }    void fun() {        scanf("%d%d%d", &K, &C, &M);        for (int i = 0; i < K + C; ++i) {            for (int j = 0; j < K + C; ++j) {                scanf("%d", &Ga[i][j]);                Ga[i][j] = Ga[i][j] ? Ga[i][j] : INF;            }        }        printf("%d\n", binary_search());    }}ac;int main(){#ifdef LOCAL    freopen("in.txt", "r", stdin);    //freopen("out.txt", "w", stdout);#endif    ac.fun();    return 0;}