SRM 558

对网络流的建图又有了更深一步理解

http://pan.baidu.com/share/link?shareid=1648064038&uk=1090822741

看了这个以后。

大概思路就是吧多元关系转化成两元关系，每个点多加两个收益点（一个表示放本位置收益，一个表示放周围收益），两个收益点分别连源点汇点（为了控制两个收益互斥）

然后就是二元关系，剩下建图就很常规了，也用不到上面论文中的解方程。

题解貌似是另一种方法，明天再看！

#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>#include <cstring>using namespace std;class SurroundingGame {public:int maxScore(vector<string> , vector<string> );};#define N 20009#define M 2000009#define inf 2000000000#define INT intstruct E {int t, next;INT flow, cap;} edge[M];int node[N], eid;vector<int> e[N];int rr[N];int que[N];void init() {eid = 0;memset(rr, -1, sizeof(rr));memset(node, -1, sizeof(node));}void addedge(int a, int b, INT c) {edge[eid].t = b;edge[eid].cap = c;edge[eid].flow = 0;edge[eid].next = node[a];node[a] = eid++;edge[eid].t = a;edge[eid].cap = 0; //0单向，c双向edge[eid].flow = 0;edge[eid].next = node[b];node[b] = eid++;}void addedge1(int a, int b, INT c) {edge[eid].t = b;edge[eid].cap = c;edge[eid].flow = 0;edge[eid].next = node[a];node[a] = eid++;edge[eid].t = a;edge[eid].cap = c; //0单向，c双向edge[eid].flow = 0;edge[eid].next = node[b];node[b] = eid++;}int bfs(int s, int t, int n) {memset(rr, -1, sizeof(rr));int i;for (i = 0; i < n; ++i)e[i].clear();int u, v;int front = 0, rear = 1;rr[s] = 0;que[0] = s;while (rear > front) {u = que[front++];for (i = node[u]; i != -1; i = edge[i].next) {v = edge[i].t;if (rr[v] == -1 && edge[i].cap) {que[rear++] = v;rr[v] = rr[u] + 1;}if (rr[v] == rr[u] + 1)e[u].push_back(i);}}return (rr[t] != -1);}INT dinic(int s, int t, int n) {int st[N];INT maxflow = 0;int aux[N];int top, cur;int p, i, k;while (bfs(s, t, n)) {top = 0;st[top] = s;cur = s;while (1) {if (cur == t) {INT minc = inf;for (i = 0; i < top; ++i) {p = aux[i + 1];if (minc > edge[p].cap)minc = edge[p].cap, k = i;}for (i = 0; i < top; ++i) {p = aux[i + 1];edge[p].cap -= minc;edge[p ^ 1].cap += minc;}maxflow += minc;cur = st[top = k];}int len = e[cur].size();while (len) {p = e[cur][len - 1];if (edge[p].cap && rr[edge[p].t] == rr[cur] + 1)break;else {len--;e[cur].pop_back();}}if (len) {cur = st[++top] = edge[p].t;aux[top] = p;} else {if (top == 0)break;rr[cur] = -1;cur = st[--top];}}}return maxflow;}int f(char c) {if ('0' <= c && '9' >= c)return c - '0';if ('a' <= c && 'z' >= c)return c - 'a' + 10;if ('A' <= c && 'Z' >= c)return c - 'A' + 36;}int dir[4][2] = { 0, 1, 1, 0, 0, -1, -1, 0 };int SurroundingGame::maxScore(vector<string> cost, vector<string> benefit) {int i, j, k;int ans;int n = cost.size();int m = cost[0].size();int s, t;s = 3 * n * m;t = s + 1;ans = 0;init();for (i = 0; i < n; ++i) {for (j = 0; j < m; ++j) {ans += 2 * f(benefit[i][j]);if ((i + j) % 2) {addedge(s, i * m + j, f(cost[i][j]));addedge(i * m + j + n * m, t, f(benefit[i][j]));addedge(s, i * m + j + n * m * 2, f(benefit[i][j]));addedge(i * m + j, i * m + j + n * m, inf);addedge(i * m + j + n * m * 2, i * m + j + n * m, inf);} else {addedge(i * m + j, t, f(cost[i][j]));addedge(i * m + j + n * m, t, f(benefit[i][j]));addedge(s, i * m + j + n * m * 2, f(benefit[i][j]));addedge(i * m + j + n * m * 2, i * m + j, inf);addedge(i * m + j + n * m * 2, i * m + j + n * m, inf);}}}for (i = 0; i < n; ++i) {for (j = 0; j < m; ++j) {for (k = 0; k < 4; ++k) {int ii = i + dir[k][0];int jj = j + dir[k][1];if (ii < 0 || jj < 0 || ii >= n || jj >= m)continue;if ((i + j) % 2) {addedge(i * m + j + n * m * 2, ii * m + jj, inf);} else {addedge(ii * m + jj, i * m + j + n * m, inf);}}}}return ans - dinic(s, t, t + 1);}