2010TianjinRegional 部分题解

比赛链接：http://acm.hust.edu.cn/vjudge/contest/129879#overview

A.Arranging Your Team

题意：给你23个人的姓名，分数，位置。让你从这23个人里挑选出一个4-4-2-1的阵容，每两个人之间会相互影响，取决于一个值。求一种挑选方法，使得整个分数最大。

4-4-2-1，每个人的职责不一样，这个复杂度分析起来挺玄学的。暴力搜索，分职责去挑选，最终挑选完毕统计结果即可。也可以用状压直接枚举，复杂度应该是一样的。注意这里应当把姓名离散化，否则会TLE（不离散化大概一个样例要跑5s，不要问我怎么知道的…）

/*
━━━━━┒ギリギリ♂ eye！
┓┏┓┏┓┃キリキリ♂ mind！
┛┗┛┗┛┃＼○／
┓┏┓┏┓┃ /
┛┗┛┗┛┃ノ)
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┃┃┃┃┃┃
┻┻┻┻┻┻
*/
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%I64d", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, x) memset((a), (x), sizeof(a))
#define Full(a) memset((a), 0x7f7f7f, sizeof(a))
#define lrt rt << 1
#define rrt rt << 1 | 1
#define pi 3.14159265359
#define RT return
#define lowbit(x) x & (-x)
#define onecnt(x) __builtin_popcount(x)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef pair<LL, LL> pll;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb;

const int maxn = 33;
typedef struct Node {
    int id;
    string name;
    int s;
    int t;
}Node;
Node fuck;
int n, m;
string tmp;
int ret;
vector<Node> peo[6];
int G[maxn][maxn];
vector<int> ans;
int score[maxn];
map<string, int> ns;

int get() {
    if(tmp[0] == 'g') return 1;
    else if(tmp[0] == 'd') return 2;
    else if(tmp[0] == 'm') return 3;
    else if(tmp[0] == 's') return 4;
}

void dfs1(int cur, int cnt);
void dfs2(int cur, int cnt);
void dfs3(int cur, int cnt);
void dfs4(int cur, int cnt);

void dfs1(int cur, int cnt) {
    if(cnt == 1) {
        dfs2(-1, 0);
        return;
    }
    For(i, cur+1, peo[1].size()) {
        ans.push_back(peo[1][i].id);
        dfs1(i, cnt+1);
        ans.pop_back();
    }
}

void dfs2(int cur, int cnt) {
    if(cnt == 4) {
        dfs3(-1, 0);
        return;
    }
    For(i, cur+1, peo[2].size()) {
        ans.push_back(peo[2][i].id);
        dfs2(i, cnt+1);
        ans.pop_back();
    }
}

void dfs3(int cur, int cnt) {
    if(cnt == 4) {
        dfs4(-1, 0);
        return;
    }
    For(i, cur+1, peo[3].size()) {
        ans.push_back(peo[3][i].id);
        dfs3(i, cnt+1);
        ans.pop_back();
    }
}

void dfs4(int cur, int cnt) {
    if(cnt == 2) {
        int sum = 0;
        Rep(i, ans.size()) {
            For(j, i+1, ans.size()) {
                sum += G[ans[i]][ans[j]];
            }
            sum += score[ans[i]];
        }
        ret = max(ret, sum);
        return;
    }
    For(i, cur+1, peo[4].size()) {
        ans.push_back(peo[4][i].id);
        dfs4(i, cnt+1);
        ans.pop_back();
    }
}

int main() {
    // FRead();
    int id = 1;
    while(cin >> fuck.name >> fuck.s >> tmp) {
        ret = -0x7f7f7f; Cls(score); Cls(G);
        For(i, 1, 5) peo[i].clear(); id = 1; ns.clear(); ans.clear();

        fuck.t = get(); ns[fuck.name] = id++;
        score[id-1] = fuck.s;
        fuck.id = id - 1;
        peo[fuck.t].pb(fuck);
        For(i, 2, 24) {
            cin >> fuck.name >> fuck.s >> tmp;
            fuck.t = get(); ns[fuck.name] = id++;
            score[id-1] = fuck.s;
            fuck.id = id - 1;
            peo[fuck.t].pb(fuck);
        }
        Rint(m);
        string a, b;
        int sc;
        Rep(i, m) {
            cin >> a >> b >> sc;
            G[ns[a]][ns[b]] = G[ns[b]][ns[a]] = sc;
        }
        if(peo[1].size()<1||peo[2].size()<4||peo[3].size()<4||peo[4].size()<2) {
            printf("impossible
");
            continue;
        }
        dfs1(-1, 0);
        printf("%d
", ret);
    }
    RT 0;
}

C.Card Game

题意：给n个字符串，让它们两两配对，比如字符串i和j，让i和j的翻转重叠的最多。

根据i,j的重叠字符数量建图G(i,j)，跑KM。模版题，注意别忘了初始化nx, ny…

/*
━━━━━┒ギリギリ♂ eye！
┓┏┓┏┓┃キリキリ♂ mind！
┛┗┛┗┛┃＼○／
┓┏┓┏┓┃ /
┛┗┛┗┛┃ノ)
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┃┃┃┃┃┃
┻┻┻┻┻┻
*/
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%I64d", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, x) memset((a), (x), sizeof(a))
#define Full(a) memset((a), 0x7f7f7f, sizeof(a))
#define lrt rt << 1
#define rrt rt << 1 | 1
#define pi 3.14159265359
#define RT return
#define lowbit(x) x & (-x)
#define onecnt(x) __builtin_popcount(x)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef pair<LL, LL> pll;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb;

/*  KM算法
 *   复杂度O(nx*nx*ny)
 *  求最大权匹配
 *   若求最小权匹配，可将权值取相反数，结果取相反数
 *  点的编号从0开始
 */
const int maxn = 310;
const int inf = 0x3f3f3f3f;
int nx,ny;//两边的点数，注意修改
int G[maxn][maxn];//二分图描述
int linker[maxn],lx[maxn],ly[maxn];//y中各点匹配状态，x,y中的点标号
int slack[maxn];
bool visx[maxn],visy[maxn];

bool dfs(int x) {
    visx[x] = true;
    for(int y = 0; y < ny; y++) {
        if(visy[y])continue;
        int tmp = lx[x] + ly[y] - G[x][y];
        if(tmp == 0) {
            visy[y] = true;
            if(linker[y] == -1 || dfs(linker[y])) {
                linker[y] = x;
                return true;
            }
        }
        else if(slack[y] > tmp)
            slack[y] = tmp;
    }
    return false;
}
int km() {
    memset(linker,-1,sizeof(linker));
    memset(ly,0,sizeof(ly));
    for(int i = 0;i < nx;i++) {
        lx[i] = -inf;
        for(int j = 0;j < ny;j++)
            if(G[i][j] > lx[i])
                lx[i] = G[i][j];
    }
    for(int x = 0;x < nx;x++) {
        for(int i = 0;i < ny;i++)
            slack[i] = inf;
        while(true) {
            memset(visx,false,sizeof(visx));
            memset(visy,false,sizeof(visy));
            if(dfs(x))break;
            int d = inf;
            for(int i = 0;i < ny;i++)
                if(!visy[i] && d > slack[i])
                    d = slack[i];
            for(int i = 0;i < nx;i++)
                if(visx[i])
                    lx[i] -= d;
            for(int i = 0;i < ny;i++) {
                if(visy[i])ly[i] += d;
                else slack[i] -= d;
            }
        }
    }
    int res = 0;
    for(int i = 0;i < ny;i++)
        if(linker[i] != -1)
            res += G[linker[i]][i];
    return res;
}

int n;
string s[maxn];

int get(string a, string b) {
    int ret = 0;
    int i, j;
    for(i = a.length() - 1, j = 0; a[i] && b[j]; i--, j++) {
        if(a[i] != b[j]) return ret;
        else ret++;
    }
    return ret;
}

int main() {
    // FRead();
    while(~Rint(n)) {
        Cls(G);
        nx = ny = n;
        Rep(i, n) cin >> s[i];
        Rep(i, n) {
            Rep(j, n) {
                if(i == j) continue;
                G[i][j] = get(s[i], s[j]);
            }
        }
        printf("%d
", km());
    }
    RT 0;
}

D.Delta Wave

题意：在一个网格上，若限定每步只能向右移动一格，可以右上，右下，横向，向右，并禁止移动到以下的地方，则以这种走法移动步从到的可能形成的路径的总数为的默慈金数。

大数+默慈金数，直接搬了http://blog.csdn.net/acdreamers/article/details/41213667

//默慈金数
// Mn = ∑C(n, 2*i) Catalan(i)
import java.math.*;
import java.util.*;

public class Main {
    
    public static final int N = 10005;
    public static final BigInteger MOD = BigInteger.valueOf(10).pow(100);
    public static BigInteger ans[] = new BigInteger[N];
    public static void Init(){
        ans[1] = BigInteger.valueOf(1);
        ans[2] = BigInteger.valueOf(2);
        for(int i = 3; i < N; i++){
            BigInteger a = ans[i - 1].multiply((BigInteger.valueOf(2).multiply(BigInteger.valueOf(i)).add(BigInteger.valueOf(1))));
            BigInteger b = ans[i - 2].multiply((BigInteger.valueOf(3).multiply(BigInteger.valueOf(i)).subtract(BigInteger.valueOf(3))));
            ans[i] = (a.add(b)).divide(BigInteger.valueOf(i).add((BigInteger.valueOf(2))));
        }
    }
    
    public static void main(String[] args){
        Init();
        Scanner cin = new Scanner(System.in);
        while(cin.hasNext()){
            int n = cin.nextInt();
            System.out.println(ans[n].mod(MOD));
        }
    }
}

E.Encoded Barcodes

题意：给一个字典和m个条形码，条形码的条形有粗有细，在一个阈值外是粗，阈值内是细。把条形码转换成字符串后是字典中的多少个字符串的前缀，统计m个条形码的总和。

字典树，条形码区分01的时候是和平均数做的比较，比平均数大是1，小是0。不能用cin，会TLE。

/*
━━━━━┒ギリギリ♂ eye！
┓┏┓┏┓┃キリキリ♂ mind！
┛┗┛┗┛┃＼○／
┓┏┓┏┓┃ /
┛┗┛┗┛┃ノ)
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┃┃┃┃┃┃
┻┻┻┻┻┻
*/
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%I64d", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, x) memset((a), (x), sizeof(a))
#define Full(a) memset((a), 0x7f7f7f, sizeof(a))
#define lrt rt << 1
#define rrt rt << 1 | 1
#define pi 3.14159265359
#define RT return
#define lowbit(x) x & (-x)
#define onecnt(x) __builtin_popcount(x)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef pair<LL, LL> pll;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb;

typedef struct Node {
    int cnt;
    Node* next[26];
    Node() {
        cnt = 0;
        Rep(i, 26) next[i] = NULL;
    }
}Node;

Node* rt;
Node memory[110000];
int mcnt;

void insert(Node* p, string str) {
    for(int i = 0; str[i]; i++) {
        if(p->next[str[i]-'a'] == NULL) {
            p->next[str[i]-'a'] = &memory[mcnt++];
        }
        p = p->next[str[i]-'a'];
        p->cnt++;
    }
}

int find(Node* p, string str) {
    for(int i = 0; str[i]; i++) {
        int t = str[i] - 'a';
        if(p->next[t] == NULL) return 0;
        p = p->next[t];
    }
    return p->cnt;
}

int n, m, k;

inline bool scan_lf(double &num) {
    char in;double Dec=0.1;
    bool IsN=false,IsD=false;
    in=getchar();
    if(in==EOF) return false;
    while(in!='-'&&in!='.'&&(in<'0'||in>'9'))
            in=getchar();
    if(in=='-'){IsN=true;num=0;}
    else if(in=='.'){IsD=true;num=0;}
    else num=in-'0';
    if(!IsD){
            while(in=getchar(),in>='0'&&in<='9'){
                    num*=10;num+=in-'0';}
    }
    if(in!='.'){
            if(IsN) num=-num;
            return true;
    }else{
            while(in=getchar(),in>='0'&&in<='9'){
                    num+=Dec*(in-'0');Dec*=0.1;
            }
    }
    if(IsN) num=-num;
    return true;
}

string get(int k) {
    string ret;
    Rep(i, k) {
        double a[10];
        double sum = 0.0;
        Rep(j, 8) {
            scan_lf(a[j]);
            sum += a[j];
        }
        sum /= 8;
        int ch = 0;
        Rep(j, 8) {
            ch <<= 1;
            ch |= (a[j] > sum);
        }
        ret.push_back(ch);
    }
    return ret;
}

int main() {
    // FRead();
    string tmp;
    while(cin >> n >> m) {
        mcnt = 0; Cls(memory);
        rt = &memory[mcnt++];
        // rt = new Node();
        Rep(i, n) {
            cin >> tmp;
            insert(rt, tmp);
        }
        int ret = 0;
        W(m) {
            Rint(k);
            tmp = get(k);
            ret += find(rt, tmp);
        }
        printf("%d
", ret);
    }
    RT 0;
}

I.I'm Telling the Truth

题意：n个人说自己的名次区间，其中有人是说谎的。求说谎人最少的情况，输出说真话的人数，要反字典序（最大的几个）。

贪心无法得到规定的序列，应该按区间的左右端点建二分图，跑匈牙利算法。

/*
━━━━━┒ギリギリ♂ eye！
┓┏┓┏┓┃キリキリ♂ mind！
┛┗┛┗┛┃＼○／
┓┏┓┏┓┃ /
┛┗┛┗┛┃ノ)
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┃┃┃┃┃┃
┻┻┻┻┻┻
*/
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, x) memset((a), (x), sizeof(a))
#define Full(a) memset((a), 0x7f7f7f, sizeof(a))
#define lrt rt << 1
#define rrt rt << 1 | 1
#define pi 3.14159265359
#define RT return
#define lowbit(x) x & (-x)
#define onenum(x) __builtin_popcount(x)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef pair<LL, LL> pll;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb;

const int maxn = 110;
const int maxm = 200100;
int x[maxn], y[maxn], linker[maxm];
int n, ret;
int ok[maxn];
bool vis[maxm];

bool dfs(int u) {
    For(i, x[u], y[u]+1) {
        if(!vis[i]) {
            vis[i] = 1;
            if(linker[i] == -1 || dfs(linker[i])) {
                linker[i] = u;
                ok[u] = i;
                return 1;
            }
        }
    }
    return 0;
}

int main() {
    // FRead();
    int T;
    Rint(T);
    W(T) {
        Rint(n);
        Cls(ok); Clr(linker, -1); Cls(vis); ret = 0;
        For(i, 1, n+1) {
            Rint(x[i]); Rint(y[i]);
        }
        for(int i = n; i >= 1; i--) {
            Cls(vis);
            if(dfs(i)) ret++;
        }
        printf("%d
", ret);
        vi fuck;
        For(i, 1, n+1) {
            if(ok[i]) fuck.pb(i);
        }
        printf("%d", fuck[0]);
        For(i, 1, fuck.size()) printf(" %d", fuck[i]);
        printf("
");
    }
    RT 0;
}