HDU 3820 Golden Eggs（最大独立集）

题目地址
题意：你有n*m个格子，你可以选择往里面放金蛋或者是银蛋或者不放，放进去就会有价值的增加，但是如果你往相邻的格子里面放相同类型的蛋，如果是金蛋则花费G的代价，如果是银蛋则花费S的代价，求最大的价值是多少?
思路：奇点(i+j)%2==1的格子的第一种类型和源点S建边，第二种类型和汇点T建边，容量为该类型在该格子对应的权值。偶点(i+j)%2==0的格子的第一种类型和汇点T建边，第二种类型和源点S建边，容量为该类型在该格子对应的权值。这样由于同一个格子我们只能取走其中一个类型的权值，那么对这两个类型建边无穷大约束一下就好，这样就代表只能选这个蛋放。然后对每个点和源点相连的类型对相邻点的同一类型建边就行了，容量就是要花的代价。（记得要把点分割开来）
总结的位置希望大家赏脸(〃’▽’〃)
吐槽：不知道为什么Dinic是TLE但是Sap是AC而且才93ms，我想应该不会相差那么大吧？求解

#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 200010#define M 800010#define LL __int64#define inf 0x3f3f3f3f#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];int mapp1[60][60], mapp2[60][60];int n, m, cnt, ans;struct node {    int to;    int cap;//剩余流量    int next;}edge[2 * M];int dir[4][2] = { { 1,0 },{ 0,1 },{ -1,0 },{ 0,-1 } };bool check(int x, int y) {// 判断是否越界    if (x>0 && x <= n && y>0 && y <= m)        return true;    return false;}bool vis[N];int len[N];int gap[N];struct Sap {    void init() {        memset(head, -1, sizeof(head));        cnt = 0;    }    void add(int u, int v, int 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    }    int dfs(int u, int s, int t, int num) {        if (u == t) {            return num;        }        int v, f, minlen = ans - 1, flow = 0;        for (int i = head[u]; i != -1; i = edge[i].next) {            v = edge[i].to;            if (edge[i].cap <= 0) continue;            if (len[v] + 1 == len[u]) {                f = dfs(v, s, t, min(edge[i].cap, num - flow));                edge[i].cap -= f;                edge[i ^ 1].cap += f;                flow += f;                if (flow == num) {                    break;                }                if (len[s] >= ans) return flow;            }            minlen = min(len[v], minlen);        }        if (flow == 0) {            if (--gap[len[u]] == 0) {                len[s] = ans;            }            len[u] = minlen + 1;            gap[len[u]]++;        }        return flow;    }    int minflow(int s, int t) {        int sum = 0;        memset(len, 0, sizeof(len));        memset(gap, 0, sizeof(gap));        gap[0] = ans;        while (len[s] < ans) {            sum += dfs(s, s, t, inf);        }        return sum;    }}sap;int main() {    cin.sync_with_stdio(false);    int G, S;    int sum;    int T, Case = 1;    cin >> T;    while (T--) {        cin >> n >> m >> G >> S;        sap.init();        sum = 0;        for (int i = 1; i <= n; i++) {            for (int j = 1; j <= m; j++) {                cin >> mapp1[i][j];                sum += mapp1[i][j];            }        }        for (int i = 1; i <= n; i++) {            for (int j = 1; j <= m; j++) {                cin >> mapp2[i][j];                sum += mapp2[i][j];            }        }        int a, b;        int s = 0, t = 2*n*m + 1;        for (int i = 1; i <= n; i++) {//奇偶建点法            for (int j = 1; j <= m; j++) {                int pos = (i - 1)*m + j;                if ((i + j) % 2 == 0) {//pos第一类型，pos+n*m第二类型                    sap.add(s, pos, mapp1[i][j]);                    sap.add(pos + n*m, t, mapp2[i][j]);                    sap.add(pos, pos + n*m, inf);                }                else {                    sap.add(s, pos, mapp2[i][j]);                    sap.add(pos + n*m, t, mapp1[i][j]);                    sap.add(pos, pos + n*m, inf);                }            }        }        for (int i = 1; i <= n; i++) {            for (int j = 1; j <= m; j++) {                int pos = (i - 1)*m + j;                for (int k = 0; k < 4; k++) {                    int x = i + dir[k][0];                    int y = j + dir[k][1];                    if (check(x, y) && (i + j) % 2 == 0) {                        sap.add(pos, (x - 1)*m + y + n*m, G);//因为奇偶点的类型是反向的                    }                    else if(check(x, y) && (i + j) % 2 == 1){                        sap.add(pos, (x - 1)*m + y + n*m, S);                    }                }            }        }        ans = 2 * n*m + 2;        cout << "Case " << Case++ << ": " << sum - sap.minflow(s, t) << endl;    }    return 0;}