LightOJ - 1198 Karate Competition(带权二分图)

题目大意：你有一个队伍，对手也有一个队伍，每个队伍里面有N个人，每个人有相应的能力值。现在要求进行N场比赛，每个人只能参加一次，如果赢一场的话，得2分，平局得1分，输了得0分，问最多能赢多少分

解题思路：模版题了

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXNODE = 510;typedef int Type;const Type INF = 0x3f3f3f3f; //1e20struct KM{    int n, m;    //邻接矩阵存储边    Type g[MAXNODE][MAXNODE];    Type Lx[MAXNODE], Ly[MAXNODE], slack[MAXNODE];    //left[i]表示Y集的i所匹配的X集的位置，right[i]表示的是X集的i匹配的Y集的位置    int left[MAXNODE], right[MAXNODE];    bool S[MAXNODE], T[MAXNODE];    void init(int n, int m) {        this->n = n;        this->m = m;        memset(g, 0, sizeof(g));    }    void AddEdge(int u, int v, Type val) {        g[u][v] += val;    }    bool dfs(int u) {        S[u] = true;        for (int v = 0; v < m; v++) {            //如果已经在交错树中出现过了，更新的值也不会变            if (T[v]) continue;            Type tmp = Lx[u] + Ly[v] - g[u][v];            if (!tmp) {                T[v] = true;                //如果该点还没有被匹配到，表示找到了一条增广链了，反之，继续增广下去                if (left[v] == -1 || dfs(left[v])) {                    left[v] = u;                    right[u] = v;                    return true;                }            }            else slack[v] = min(slack[v], tmp);        }        return false;    }    void update() {        Type a = INF;        //找到Y集合的不在交错树上的点        for (int i = 0; i < m; i++)            if (!T[i]) a = min(a, slack[i]);        for (int i = 0; i < n; i++)            if (S[i]) Lx[i] -= a;        for (int j = 0; j < m; j++)            if (T[j]) Ly[j] += a;    }    Type km() {        memset(left, -1 ,sizeof(left));        memset(right, -1, sizeof(right));        memset(Ly, 0, sizeof(Ly));        //初始化的时候，Lx初始化为最大的，Ly初始化为0，这样可以保证最优        for (int i = 0; i < n; i++) {            Lx[i] = -INF;            for (int j = 0; j < m; j++)                Lx[i] = max(Lx[i], g[i][j]);        }        for (int i = 0; i < n; i++) {            //初始化slack，这样就能达到N^3的复杂度了            for (int j = 0; j < m; j++) slack[j] = INF;            while (1) {                memset(S, 0, sizeof(S));                memset(T, 0, sizeof(T));                if (dfs(i)) break;                update();            }        }        Type ans = 0;        for (int i = 0; i < n; i++)            ans += g[i][right[i]];        return ans;    }}km;int n, cas = 1;int team1[MAXNODE], team2[MAXNODE];void init() {    scanf("%d", &n);    for (int i = 1; i <= n; i++)        scanf("%d", &team1[i]);    for (int i = 1; i <= n; i++)        scanf("%d", &team2[i]);    km.init(n, n);    for (int i = 1; i <= n; i++)        for (int j = 1; j <= n; j++) {            if (team1[i] == team2[j]) km.AddEdge(i - 1, j - 1, 1);            else if (team1[i] > team2[j]) km.AddEdge(i - 1, j - 1, 2);            else km.AddEdge(i - 1, j - 1, 0);        }    printf("Case %d: %d\n", cas++, km.km());}int main() {    int test;    scanf("%d", &test);    while (test--) {        init();    }    return 0;}