题意：
输入一个n行m列的图
每次按字母顺序走最短路， 从一个字母走到下一个字母的过程中，只能拿走一个金子，求走完当前图中所有的字母后能拿到的金子的最大值
解析：
bfs求最短路
对于一个金子如果 dis1[i] + dis2[i] == dis1[next] 那么就代表着这个金子 在这条最短路上 可以拿 那么从上一个 字母 到当前节点连一条边 权值为1
会了吧。。。
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <cmath> #include <bitset> #define rap(i, a, n) for(int i=a; i<=n; i++) #define rep(i, a, n) for(int i=a; i<n; i++) #define lap(i, a, n) for(int i=n; i>=a; i--) #define lep(i, a, n) for(int i=n; i>a; i--) #define rd(a) scanf("%d", &a) #define rlld(a) scanf("%lld", &a) #define rc(a) scanf("%c", &a) #define rs(a) scanf("%s", a) #define pd(a) printf("%d ", a); #define plld(a) printf("%lld ", a); #define pc(a) printf("%c ", a); #define ps(a) printf("%s ", a); #define MOD 2018 #define LL long long #define ULL unsigned long long #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define _ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int maxn = 110000, INF = 0x7fffffff, LL_INF = 0x7fffffffffffffff; int n, m, s, t; int d[maxn], vis[maxn]; int head[maxn], cur[maxn], cnt, dis1[maxn], dis2[maxn]; char str[110][110]; int dis[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}}; vector<int> g; struct edge { int u, v, c, next; }Edge[maxn]; void add_(int u, int v, int c) { Edge[cnt].u = u; Edge[cnt].v = v; Edge[cnt].c = c; Edge[cnt].next = head[u]; head[u] = cnt++; } void add(int u, int v, int c) { add_(u, v, c); add_(v, u, 0); } struct node { int alpha, idx; bool operator < (const node &a) const{ return alpha < a.alpha; } }Node[maxn]; void bfs1(int s) { for(int i = 0; i < maxn; i++) d[i] = INF; queue<int> Q; mem(vis, 0); Q.push(s); vis[s] = 1; d[s] = 0; while(!Q.empty()) { int u = Q.front(); Q.pop(); for(int i = 0; i < 4; i++) { int v = u + dis[i][0] * m + dis[i][1]; if(u % m == 0 && i == 0 || (u - 1) % m == 0 && i == 1) continue; if(v < 1 || v > n * m || vis[v] || str[v / m + 1][v % m] == '#') continue; d[v] = d[u] + 1; vis[v] = 1; Q.push(v); } } } bool bfs2() { mem(d, 0); queue<int> Q; Q.push(0); d[0] = 1; while(!Q.empty()) { int u = Q.front(); Q.pop(); for(int i = head[u]; i != -1; i = Edge[i].next) { edge e = Edge[i]; if(!d[e.v] && e.c > 0) { d[e.v] = d[u] + 1; Q.push(e.v); if(e.v == t) return 1; } } } return d[t] != 0; } int dfs(int u, int cap) { int ret = 0; if(u == t || cap == 0) return cap; for(int &i = cur[u]; i != -1; i = Edge[i].next) { edge e = Edge[i]; if(d[e.v] == d[u] + 1 && e.c > 0) { int V = dfs(e.v, min(e.c, cap)); Edge[i].c -= V; Edge[i ^ 1].c += V; ret += V; cap -= V; if(cap == 0) break; } } if(cap > 0) d[u] = -1; return ret; } int Dinic() { mem(d, 0); int ans = 0; while(bfs2()) { memcpy(cur, head, sizeof(head)); ans += dfs(0, INF); } return ans; } int main() { while(~scanf("%d%d", &n, &m)) { g.clear(); t = 10500; mem(head, -1); cnt = 0; int ans = 0; for(int i = 1; i <= n; i++) { rs(str[i] + 1); for(int j = 1; j <= m; j++) { if(str[i][j] >= 'A' && str[i][j] <= 'z') Node[ans].alpha = str[i][j], Node[ans++].idx = (i - 1) * m + j; if(str[i][j] == '*') g.push_back((i - 1) * m + j), add((i - 1) * m + j + 100, 10500, 1); } } sort(Node, Node + ans); int num = 0, flag = 0; for(int i = 0; i < ans - 1; i++) { ++num; bfs1(Node[i].idx); memcpy(dis1, d, sizeof(d)); if(dis1[Node[i + 1].idx] == INF) { flag = 1; break; } bfs1(Node[i + 1].idx); memcpy(dis2, d, sizeof(d)); for(int j = 0; j < g.size(); j++) if(dis1[g[j]] + dis2[g[j]] == dis1[Node[i + 1].idx]) add(num, g[j] + 100, 1); } if(flag == 1) { cout << "-1" << endl; continue; } for(int i = 1; i <= 52; i++) add(0, i, 1); cout << Dinic() << endl; } return 0; }