HDU - 3035 War(对偶图+最小割+最短路)
2015-09-07 15:29
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题目大意:给出一个方格,要求从(0,0)位置运送尽量多的货物到(N,M),给出每条边的容量,问最多可以运送多少过去
解题思路:这题要转换成对偶图去做,然后求最短路,如果用最大流的话,光看边和点的数量就足以爆炸了
给出对偶图的讲解浅析最大最小定理在信息学竞赛中的应用
解题思路:这题要转换成对偶图去做,然后求最短路,如果用最大流的话,光看边和点的数量就足以爆炸了
给出对偶图的讲解浅析最大最小定理在信息学竞赛中的应用
[code]#include <cstdio> #include <cstring> #include <algorithm> #include <queue> using namespace std; const int MAXNODE = 1000010; const int MAXEDGE = MAXNODE * 4 * 2; const int INF = 0x3fffffff; struct Edge{ int v, val, next; Edge() {} Edge(int v, int val, int next): v(v), val(val), next(next) {} }E[MAXEDGE]; int head[MAXNODE], d[MAXNODE]; int n, m, source, sink, tot; bool vis[MAXNODE]; void AddEdge(int u, int v, int val) { E[tot] = Edge(v, val, head[u]); head[u] = tot++; u = u ^ v; v = u ^ v; u = u ^ v; E[tot] = Edge(v, val, head[u]); head[u] = tot++; } void init() { memset(head, -1, sizeof(head)); tot = 0; source = n * m * 4 + 1, sink = source + 1; int u, v, val; for (int i = 0; i < m; i++) { scanf("%d", &val); u = i * 4 + 2; AddEdge(sink, u, val); } for (int i = 1; i < n; i++) { for (int j = 0; j < m; j++) { scanf("%d", &val); u = 4 + (i - 1) * m * 4 + j * 4; v = 2 + i * m * 4 + j * 4; AddEdge(u, v, val); } } for (int i = 1; i <= m; i++) { scanf("%d", &val); u = (n - 1) * m * 4 + i * 4; AddEdge(u, source, val); } for (int i = 0; i < n; i++) { scanf("%d", &val); AddEdge(source, 1 + i * m * 4, val); for (int j = 1; j < m; j++) { scanf("%d", &val); u = i * m * 4 + 3 + (j - 1) * 4; v = u + 2; AddEdge(u, v, val); } scanf("%d", &val); AddEdge((i + 1) * m * 4 - 1, sink, val); } for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { scanf("%d", &val); u = 1 + i * m * 4 + j * 4; AddEdge(u, u + 1, val); scanf("%d", &val); AddEdge(u + 1, u + 2, val); } for (int j = 0; j < m; j++) { scanf("%d", &val); u = 1 + i * m * 4 + j * 4; AddEdge(u, u + 3, val); scanf("%d", &val); AddEdge(u + 3, u + 2, val); } } } struct Node { int v, val; Node() {} Node(int v, int val): v(v), val(val) {} bool operator < (const Node &a) const { return a.val < val; } }; void Dijkstra() { for (int i = 1; i <= sink; i++) { d[i] = INF; vis[i] =false; } d[source] = 0; priority_queue<Node> Q; Q.push(Node(source, 0)); while (!Q.empty()) { Node t = Q.top(); Q.pop(); if (t.v == sink) break; if (vis[t.v]) continue; vis[t.v] = true; for (int i = head[t.v]; ~i; i = E[i].next) { int v = E[i].v; if (!vis[v] && d[v] > d[t.v] + E[i].val) { d[v] = d[t.v] + E[i].val; Q.push(Node(v, d[v])); } } } printf("%d\n", d[sink]); } int main() { while (scanf("%d%d", &n, &m) != EOF) { init(); Dijkstra(); } return 0; }
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