【HDOJ】3505 Writing Robot

挺好的一道题目，我的做法是kmp+Dinic网络流。
kmp求子串在P中出现的次数，从而计算love值。
网络流主要用来处理最优解。
case2中p1的love值是8，p2的love值是7，最终T包含p1和p2，hate值也仅仅算一次。
这个题目难点在于思考为什么网络流的解法是合理，可以反证。从而导出最优解等于love的总和-最大流。
建图方法：
source连接p，初始容量为love值；
p连接s，初始容量为INF;
s连接destination，容量为hate值。
在最优解中，如果s到des的流量小于容量，则证明包含s的p都不被选择。即减去p的容量和。如果流量大于等于容量，则证明该直接去掉hate值。

/* 3505 */
#include <iostream>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <deque>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <climits>
#include <cctype>
#include <cassert>
#include <functional>
#include <iterator>
#include <iomanip>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,1024000")

#define sti                set<int>
#define stpii            set<pair<int, int> >
#define mpii            map<int,int>
#define vi                vector<int>
#define pii                pair<int,int>
#define vpii            vector<pair<int,int> >
#define rep(i, a, n)     for (int i=a;i<n;++i)
#define per(i, a, n)     for (int i=n-1;i>=a;--i)
#define clr                clear
#define pb                 push_back
#define mp                 make_pair
#define fir                first
#define sec                second
#define all(x)             (x).begin(),(x).end()
#define SZ(x)             ((int)(x).size())
#define lson            l, mid, rt<<1
#define rson            mid+1, r, rt<<1|1

typedef struct {
    int l, h, len;
    char s[54];
    int nxt[54];
    
    void getnext() {
        int i = 0, j = -1;
        
        nxt[0] = -1;
        while (i < len) {
            if (j==-1 || s[i]==s[j]) {
                ++i;
                ++j;
                nxt[i] = j;
            } else {
                j = nxt[j];
            }
        }
    }
    
    int kmp(char *d) {
        int dlen = strlen(d);
        int i = 0, j =0;
        int ret = 0;
        
        while (i < dlen) {
            if (d[i] == s[j]) {
                ++i;
                ++j;
            } else {
                j = nxt[j];
                if (j == -1) {
                    j = 0;
                    ++i;
                }
            }
            if (j == len) {
                ++ret;
                j = nxt[j];
            }
        }
        
        return ret;
    }
    
} node_t;

typedef struct {
    int v, f, nxt;
} edge_t;

const int INF = 0x3f3f3f3f;
const int maxn = 155;
const int maxv = maxn * 2;
const int maxl = 1005;
const int maxe = 1e5+5;
node_t nd[maxn];
bool M[maxn][maxn];
int score[maxn];
int head[maxv], head_[maxv];
edge_t E[maxe];
int dis[maxv];
int Q[maxv];
char s[maxl];
int nxt[maxl];
int sn, pn, m;

int calc(int k) {
    int ret = 0, cnt;
    
    rep(i, 1, sn+1) {
        cnt = nd[i].kmp(s);
        if (cnt) {
            ret += nd[i].l * cnt;
            M[k][i] = true;
        }
    }
    
    return ret;
}

void init() {
    m = 0;
    memset(head, -1, sizeof(head));
    memset(M, false, sizeof(M));
}

void addEdge(int u, int v, int f) {
    E[m].v = v;
    E[m].f = f;
    E[m].nxt = head[u];
    head[u] = m++;
    
    E[m].v = u;
    E[m].f = 0;
    E[m].nxt = head[v];
    head[v] = m++;
}

bool bfs(int s, int t) {
    int l = 0, r = 0;
    int u, v, k;
    
    Q[r++] = s;
    memset(dis, 0, sizeof(dis));
    dis[s] = 1;
    
    while (l < r) {
        u = Q[l++];
        for (k=head[u]; k!=-1; k=E[k].nxt) {
            v = E[k].v;
            if (!dis[v] && E[k].f) {
                dis[v] = dis[u] + 1;
                if (v == t)
                    return false;
                Q[r++] = v;
            }
        }
    }
    
    return true;
}

int dfs(int u, int t, int val) {
    if (u==t || val==0)
        return val;
    
    int ret = 0;
    int v, tmp;
    
    for (int& k=head_[u]; k!=-1; k=E[k].nxt) {
        v = E[k].v;
        if (E[k].f && dis[v]==dis[u]+1 && (tmp=dfs(v, t, min(val, E[k].f)))>0) {
            E[k].f -= tmp;
            E[k^1].f += tmp;
            ret += tmp;
            val -= tmp;
            if (val == 0)
                return ret;
        }
    }
    
    return ret;
}

int Dinic(int s, int t) {
    int ret = 0, tmp;
    
    while (1) {
        if (bfs(s, t))
            break;
        
        memcpy(head_, head, sizeof(head));
        while (1) {
            tmp = dfs(s, t, INF);
            if (tmp == 0)
                break;
            ret += tmp;
        }
    }
    
    return ret;
}

void Build() {
    int s = 0;
    int t = maxv - 1;
    
    rep(i, 1, sn+1) {
        addEdge(i, t, nd[i].h);
    }
    
    rep(i, 1, pn+1) {
        addEdge(s, sn+i, score[i]);
        rep(j, 1, sn+1) {
            if (M[i][j])
                addEdge(sn+i, j, INF);
        }
    }
}

int solve() {
    int ret = 0;
    
    rep(i, 1, pn+1) {
        scanf("%s", s);
        score[i] = calc(i);
        ret += score[i];
    }
    
    Build();
    int tmp = Dinic(0, maxv-1);
    ret -= tmp;
    #ifndef ONLINE_JUDGE
        printf("maxflow = %d
", tmp);
    #endif
    
    return ret;
}

int main() {
    ios::sync_with_stdio(false);
    #ifndef ONLINE_JUDGE
        freopen("data.in", "r", stdin);
        freopen("data.out", "w", stdout);
    #endif
    
    int t;
    int ans;
    
    scanf("%d", &t);
    rep(tt, 1, t+1) {
        init();
        scanf("%d %d", &sn, &pn);
        rep(i, 1, sn+1) {
            scanf("%d %d %s", &nd[i].l, &nd[i].h, nd[i].s);
            nd[i].len = strlen(nd[i].s);
            nd[i].getnext();
        }
        ans = solve();
        printf("Case %d: %d
", tt, ans);
    }
    
    #ifndef ONLINE_JUDGE
        printf("time = %d.
", (int)clock());
    #endif
    
    return 0;
}