题意
有$n?$个小朋友，给每个人分$1~m?$个糖果,有k个限制 限制形如$(x,y,z)?$ 表示第$x?$个人分到的糖数减去第$y?$个人分到的糖数不大于$z?$,给第$i?$个人$j?$颗糖获得的满意度为$w_{i,j}?$，问总满意度最大值
点$(i,j)$表示第$i$个人分$j$个糖,当这个点属于$s$集合成立，因为是求满意度最大值，所以负权建边，同时加上个最大值$Max$使得满足最大流模板，假设不考虑限制，对于每一个$i$，连边$(i,j)\rightarrow (i,j+1),j\in[1,m)$，边权为$Max-w_i,j$，$s\rightarrow(i,1)$，那么此时的最小割便是能得到最大满意度
对于每个限制,当$x$选了$i$个糖，那么$y$至少要选$i-z$个糖，连边$(x,i)\rightarrow s, i \in [1,m] \land i - z < 1$，或$(x,i)\rightarrow (y,i-z), i \land [1,m] \land 1 \le i - z \le m$ ，或$(x,i)\rightarrow t, i \in [1,m] \land i - z > m$ ，边权为$inf$，这样，当不满足限制的割边发生时，得到的最小割会大于$inf$
关于限制的建边的详细题解：http://blog.csdn.net/wing_wuchen/article/details/77407413
答案为$Max*n-mincut$
代码
#include <bits/stdc++.h>
#define MAXN 300005
#define MAXM 50000005
#define inf 0x3f3f3f3f
using namespace std;
typedef long long LL;
struct Edge {
int to, nxt, c;
}edge[MAXM];
int head[MAXN], cnt = 0;
int d[MAXN], cur[MAXN], pre[MAXN], gap[MAXN];
int source, sink, limit;
void init() {
memset(head, -1, sizeof(head));
cnt = 0;
}
inline void add_edge(int u, int v, int c) {
edge[cnt].to = v;
edge[cnt].nxt = head[u];
edge[cnt].c = c;
head[u] = cnt++;
}
inline void add(int u, int v, int c) {
add_edge(u, v, c); add_edge(v, u, 0);
}
void rev_bfs() {
memset(gap, 0, sizeof(gap));
memset(d, -1, sizeof(d));
d[sink] = 0;
gap[0] = 1;
queue<int> que;
que.push(sink);
while(!que.empty()) {
int u = que.front(); que.pop();
for(int i = head[u]; ~i; i = edge[i].nxt) {
int v = edge[i].to;
if(~d[v])continue;
d[v] = d[u] + 1;
gap[d[v]]++;
que.push(v);
}
}
}
int isap() {
memcpy(cur, head, sizeof(cur));
rev_bfs();
int flow = 0, i;
int u = source;
pre[source] = source;
while(d[sink] < limit) {
if(u == sink) {
int f = inf, neck;
for(i = source; i != sink; i = edge[cur[i]].to) {
if(f > edge[cur[i]].c) {
f = edge[cur[i]].c;
neck = i;
}
}
for(i = source; i != sink; i = edge[cur[i]].to) {
edge[cur[i]].c -= f;
edge[cur[i] ^ 1].c += f;
}
flow += f;
u = neck;
}
for(i = cur[u]; ~i; i = edge[i].nxt) {
if(d[edge[i].to] + 1 == d[u] && edge[i].c) break;
}
if(~i) {
cur[u] = i;
pre[edge[i].to] = u;
u = edge[i].to;
}else {
if((--gap[d[u]]) == 0) break;
int mind = limit;
for(int i = head[u]; ~i; i = edge[i].nxt) {
if(edge[i].c && mind > d[edge[i].to]) {
cur[u] = i;
mind = d[edge[i].to];
}
}
d[u] = mind + 1;
gap[d[u]]++;
u = pre[u];
}
}
return flow;
}
int t, n, m, k, w[100][100], x, y, z;
int get(int x, int y) {return (x - 1) * m + y;}
int main() {
scanf("%d", &t);
while(t--) {
init();
scanf("%d%d%d", &n, &m, &k);
source = 0; sink = n * m + 1; limit = sink + 1;
for(int i = 1; i <= n; ++i) {
add(source, get(i, 1), inf);
for(int j = 1; j <= m; ++j) {
scanf("%d", &w[i][j]);
if(j < m) add(get(i, j), get(i, j + 1), 1000 - w[i][j]);
else add(get(i, j), sink, 1000 - w[i][j]);
}
}
for(int i = 1; i <= k; ++i) {
scanf("%d%d%d", &x, &y, &z);
for(int j = 1; j <= m; ++j) {
if(j - z < 1) add(get(x, j), source, inf);
else if(j - z <= m) add(get(x, j), get(y, j - z), inf);
else add(get(x, j), sink, inf);
}
}
int ans = isap();
if(ans >= inf) printf("-1\n"); else printf("%d\n", 1000 * n - ans);
}
return 0;
}?
