HDU 5772 String problem（一个诡异的建图）

题目地址
题意：给你一个长度为n的只有0~9字符的字符串，要你挑出一些字符来组成一个新的字符串，然后挑选每个字符的代价是

然后呢，能得到的价值是（substr是新串，id[i]是原串的位置）就是求和，规则看下图。

思路：这个我还是搞得不太清楚，看到别人发的博客中放了官方题解，我就看了下，然后慢慢理解，但是建图就十分经典了，就是要理解点的划分的问题，官方题解讲的很详细了，我就不多说了。

#include <iostream>#include <cstring>#include <string>#include <queue>#include <vector>#include <map>#include <set>#include <stack>#include <cmath>#include <cstdio>#include <algorithm>#include <iomanip>#define N 6100#define M 100010  #define LL __int64#define inf 0x3f3f3f3f3f3f3f3f#define lson l,mid,ans<<1#define rson mid+1,r,ans<<1|1#define getMid (l+r)>>1#define movel ans<<1#define mover ans<<1|1using namespace std;const LL mod = 1000000007;int head[N], level[N], cur[N];int n, m, cnt;struct node {    int to;    LL cap;//剩余流量    int next;}edge[M];struct Dinic {    void init() {        memset(head, -1, sizeof(head));        cnt = 0;    }    void add(int u, int v, LL cap) {//有向图        edge[cnt].to = v, edge[cnt].cap = cap, edge[cnt].next = head[u], head[u] = cnt++;        edge[cnt].to = u, edge[cnt].cap = 0, edge[cnt].next = head[v], head[v] = cnt++;//反向边    }    bool bfs(int s, int t){        memset(level, -1, sizeof(level));        queue<int>Q;        level[s] = 0;        Q.push(s);        while (!Q.empty()){            int u = Q.front(); Q.pop();            for (int i = head[u]; i + 1; i = edge[i].next){                int v = edge[i].to;                if (edge[i].cap>0 && level[v] == -1){                    level[v] = level[u] + 1;                    Q.push(v);                }            }        }        return level[t] != -1;    }    LL dfs(int u, int t, LL f){        if (u == t) return f;        for (int &i = cur[u]; i + 1; i = edge[i].next){            int v = edge[i].to;            if (edge[i].cap>0 && level[v] == level[u] + 1){                LL d = dfs(v, t, min(f, edge[i].cap));                if (d>0){                    edge[i].cap -= d;                    edge[i ^ 1].cap += d;                    return d;                }            }        }        return 0;    }    LL dinic(int s, int t){        LL flow = 0;        while (bfs(s, t)){            for (int i = 0; i<N; i++) cur[i] = head[i];            LL f;            while ((f = dfs(s, t, inf))>0)                flow += f;        }        return flow;    }}dc;int main() {        cin.sync_with_stdio(false);    LL a, b, c;    int numa[10], numb[10];    int mapp[110][110];    int s, t;    LL sum;    int ans;    string str;    int T, Case = 1;    cin >> T;    while (T--) {        cin >> n;        dc.init();        sum = 0;        s = 0;        t = n*(n - 1) / 2 + n + 10 + 1;//每个由第i个点和第j个点组成的点+n个坐标+10个字符+1        cin >> str;        for (int i = 0; i < 10; i++) {            cin >> numa[i] >> numb[i];        }        for (int i = 0; i < n; i++) {            for (int j = 0; j < n; j++) {                cin >> mapp[i][j];                sum += mapp[i][j];            }        }        ans = 1;        for (int i = 0; i < n; i++) {            for (int j = i + 1; j < n; j++) {                dc.add(s, ans, mapp[i][j] + mapp[j][i]);//价值                dc.add(ans, n*(n - 1) / 2 + i + 1, inf);//n*(n - 1) / 2 + i  第几个字符                dc.add(ans, n*(n - 1) / 2 + j + 1, inf);                ans++;            }        }        for (int i = 1; i <= n; i++) {            dc.add(n*(n - 1) / 2 + i, n*(n - 1) / 2 + n + str[i - 1] - '0' + 1, inf);            dc.add(n*(n - 1) / 2 + i, t, numa[str[i - 1] - '0']);//花费        }        for (int i = 0; i < 10; i++) {            dc.add(n*(n - 1) / 2 + n + i + 1, t, (numb[i] - numa[i]));        }        cout << "Case #" << Case++ << ": " << sum - dc.dinic(s, t) << endl;    }    return 0;}