HDU 3718

来源：互联网发布：极乐净土动作镜头数据编辑：程序博客网时间：2024/06/09 16:43

颇有收获。

自己组一组连线，但是连线可以通过给相应两两字母组合提供便利。

这个输入K是卖萌的吧

那个scanf(" %c",&str[i])的手法讲道理有点意思

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int MAXNODE = 27;typedef int Type;const Type INF = 0x3f3f3f3f;struct KM {    int n, m;    Type g[MAXNODE][MAXNODE];    Type Lx[MAXNODE], Ly[MAXNODE], slack[MAXNODE];    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 add_Edge(int u, int v, Type val) {        g[u][v] += val;    }    bool dfs(int i) {        S[i] = true;        for (int j = 0; j < m; j++) {            if (T[j]) continue;            Type tmp = Lx[i] + Ly[j] - g[i][j];            if (!tmp) {                T[j] = true;                if (left[j] == -1 || dfs(left[j])) {                    left[j] = i;                    right[i] = j;                    return true;                }            } else slack[j] = min(slack[j], tmp);        }        return false;    }    void update() {        Type a = INF;        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 i = 0; i < m; i++)            if (T[i]) Ly[i] += a;    }    Type km() {        memset(left, -1, sizeof(left));        memset(right, -1, sizeof(right));        memset(Ly, 0, sizeof(Ly));        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++) {            for (int j = 0; j < m; j++) slack[j] = INF;            while (1) {                memset(S, false, sizeof(S));                memset(T, false, sizeof(T));                if (dfs(i)) break;                else update();            }        }        Type ans = 0;        for (int i = 0; i < n; i++) {            ans += g[i][right[i]];        }        return ans;    }} gao;int t,n,k,m;char str1[10000+10],str2[10000+10];int main(){    scanf("%d",&t);    while(t--){        scanf("%d%d%d",&n,&k,&m);        for(int i=0;i<n;i++){            scanf(" %c",&str1[i]);        }        int ans;        while(m--){            gao.init(26,26);            for(int i=0;i<n;i++){                scanf(" %c",&str2[i]);                gao.add_Edge(str1[i]-'A',str2[i]-'A',1);            }            ans=gao.km();            printf("%.4lf\n",ans*1.0/n);        }    }    return 0;}

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