Give out candies
题解:
第一次遇见这样处理的网络流模型。
将问题转换成最小割问题。
具体的题解参考自:传送门
先将每个人的拆成m个人。
然后s向第1人连边流量为inf。第i个人向第i+1个人连边,流量为 3000 - w。 将t视为每组的第m+1个人。
接来下是约束关系的建边, x, y ,z。
如果x小朋友拿了j个糖果,则y小朋友拿的糖果至少为y-z。
则在x的拆点和y对应的拆点之间建立一条流量为inf的边。
最后跑完最大流的时候。
如果ans >= inf 则说明没有最小割, 无解。
否则 ans = n * 3000 - ans.
代码:
![](http://img.e-com-net.com/image/info8/b8d97b5613f94ed2ba791cad57d0b2ed.gif)
![](http://img.e-com-net.com/image/info8/2f88dd3f1cd145f59c0e47b51acdbd4b.gif)
#includeusing namespace std; #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); #define LL long long #define ULL unsigned LL #define fi first #define se second #define pb push_back #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define lch(x) tr[x].son[0] #define rch(x) tr[x].son[1] #define max3(a,b,c) max(a,max(b,c)) #define min3(a,b,c) min(a,min(b,c)) typedef pair<int,int> pll; const int inf = 0x3f3f3f3f; const int _inf = 0xc0c0c0c0; const LL INF = 0x3f3f3f3f3f3f3f3f; const LL _INF = 0xc0c0c0c0c0c0c0c0; const LL mod = (int)1e9+7; const int N = 3000; const int M = 2e5; int head[N], deep[N], cur[N]; int w[M], to[M], nx[M]; int tot; void add(int u, int v, int val){ w[tot] = val; to[tot] = v; nx[tot] = head[u]; head[u] = tot++; w[tot] = 0; to[tot] = u; nx[tot] = head[v]; head[v] = tot++; } int bfs(int s, int t){ queue<int> q; memset(deep, 0, sizeof(deep)); // for(int i = s; i <= t; ++i) deep[i]=0; q.push(s); deep[s] = 1; while(!q.empty()){ int u = q.front(); q.pop(); for(int i = head[u]; ~i; i = nx[i]){ if(w[i] > 0 && deep[to[i]] == 0){ deep[to[i]] = deep[u] + 1; q.push(to[i]); } } } return deep[t] > 0; } LL Dfs(int u, int t, LL flow){ if(u == t) return flow; for(int &i = cur[u]; ~i; i = nx[i]){ if(deep[u]+1 == deep[to[i]] && w[i] > 0){ LL di = Dfs(to[i], t, min(1ll*w[i], flow)); if(di > 0){ w[i] -= di, w[i^1] += di; return di; } } } return 0; } LL Dinic(int s, int t){ LL ans = 0, tmp; while(bfs(s, t)){ for(int i = 0; i <= t; i++) cur[i] = head[i]; while(tmp = Dfs(s, t, inf)) ans += tmp; } return ans; } void init(){ memset(head, -1, sizeof(head)); tot = 0; } int main(){ int T; scanf("%d", &T); for(int _cas = 1; _cas <= T; ++_cas){ init(); int n, m, k, s, t; scanf("%d%d%d", &n, &m, &k); s = 0, t = n * m + 1; // assert(t > 300); for(int i = 0, v; i < n; ++i){ add(s, i*m+1, inf); for(int j = 1; j <= m; ++j){ scanf("%d", &v); if(j == m) add(i*m+j, t, 3000-v); else add(i*m+j, i*m+j+1, 3000-v); } } for(int i = 1, x,y,z; i <= k; ++i){ scanf("%d%d%d", &x, &y, &z); for(int j = 1; j <= m; ++j){ if(j-z < 1) add(x*m-m+j, s, inf); else if(j-z > m) add(x*m-m+j, t, inf); else add((x-1)*m+j, (y-1)*m+j-z, inf); } } LL ans = Dinic(s, t); if(ans >= inf) puts("-1"); else printf("%lld\n", 3000*n-ans); } return 0; }