POJ 3422 Kaka's Matrix Travels(模型转化 区间K覆盖问题)
2015-09-04 23:55
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题目大意:给出一个网格,每次的起点都是左上角,且只能往下或者往右走,终点是右下角,每走过一个格子,可以将该格子内的数拿走,格子内的数只能被拿走一次
问,走k次,拿到的数的总和的最大值是多少
解题思路:将其转化为区间k覆盖问题,只能走k次,那么边权最多只能是k了
我是将点拆了,一条边是容量为1,费用为该格子的数的相反数
另一条边容量是k,费用为0
然后能连的格子连起来,这样图就建成了
问,走k次,拿到的数的总和的最大值是多少
解题思路:将其转化为区间k覆盖问题,只能走k次,那么边权最多只能是k了
我是将点拆了,一条边是容量为1,费用为该格子的数的相反数
另一条边容量是k,费用为0
然后能连的格子连起来,这样图就建成了
[code]#include <cstdio> #include <cstring> #include <queue> #include <vector> #include <algorithm> using namespace std; const int MAXNODE = 10010; const int MAXEDGE = 1000010; typedef int Type; const Type INF = 0x3f3f3f3f; struct Edge{ int u, v, next; Type cap, flow, cost; Edge() {} Edge(int u, int v, Type cap, Type flow, Type cost, int next): u(u), v(v), cap(cap), flow(flow), cost(cost), next(next) {} }; struct MCMF{ int n, m, s, t; Edge edges[MAXEDGE]; int head[MAXNODE]; int p[MAXNODE]; Type d[MAXNODE]; Type a[MAXNODE]; bool inq[MAXNODE]; void init(int n) { this->n = n; memset(head, -1, sizeof(head)); m = 0; } void AddEdge(int u, int v, Type cap, Type cost) { edges[m] = Edge(u, v, cap, 0, cost, head[u]); head[u] = m++; edges[m] = Edge(v, u, 0, 0, -cost, head[v]); head[v] = m++; } bool BellmanFord(int s, int t, Type &flow, Type &cost) { for (int i = 0; i < n; i++) d[i] = INF; memset(inq, 0, sizeof(inq)); d[s] = 0; inq[s] = true; p[s] = 0; a[s] = INF; queue<int> Q; Q.push(s); while (!Q.empty()) { int u = Q.front(); Q.pop(); inq[u] = false; for (int i = head[u]; ~i; i = edges[i].next) { Edge &e = edges[i]; if (e.cap > e.flow && d[e.v] > d[u] + e.cost) { d[e.v] = d[u] + e.cost; p[e.v] = i; a[e.v] = min(a[u], e.cap - e.flow); if (!inq[e.v]) { Q.push(e.v); inq[e.v] = true; } } } } if (d[t] == INF) return false; flow += a[t]; cost += a[t] * d[t]; int u = t; while (u != s) { edges[p[u]].flow += a[t]; edges[p[u] ^ 1].flow -= a[t]; u = edges[p[u]].u; } return true; } Type MinCost(int s, int t, int k) { this->s = s; this->t = t; Type flow = 0, cost = 0; while (BellmanFord(s, t, flow, cost)){ // printf("flow is %d\n", flow); if (flow == k) break; } return cost; } }mcmf; #define maxn 60 int n, k; int num[maxn][maxn]; int dir[2][2] = {{0, -1}, {-1, 0}}; void init() { for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) scanf("%d", &num[i][j]); } void solve() { int t = n * n; mcmf.init(t * 2 + 3); for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) { mcmf.AddEdge((i - 1) * n + j, (i - 1) * n + j + t, 1, -num[i][j]); mcmf.AddEdge((i - 1) * n + j, (i - 1) * n + j + t, k, 0); } for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) for (int l = 0; l < 2; l++) { int tx = i + dir[l][0]; int ty = j + dir[l][1]; if (tx < 1 || tx > n || ty < 1 || ty > n) continue; mcmf.AddEdge((tx - 1) * n + ty + t, (i - 1) * n + j, k, 0); } mcmf.AddEdge(t * 2 + 1, 1, k, 0); mcmf.AddEdge(t * 2, t * 2 + 2, k, 0); printf("%d\n", -mcmf.MinCost(t * 2 + 1, t * 2 + 2, k)); } int main() { while (scanf("%d%d", &n, &k) != EOF) { init(); solve(); } return 0; }
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