| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 6204 | Accepted: 1389 |
Description
It was long ago when we played the game Red Alert. There is a magic function for the game objects which is called instantaneous transfer. When an object uses this magic function, it will be transferred to the specified point immediately, regardless of how far it is.
Now there is a mining area, and you are driving an ore-miner truck. Your mission is to take the maximum ores in the field.
The ore area is a rectangle region which is composed by n × m small squares, some of the squares have numbers of ores, while some do not. The ores can't be regenerated after taken.
The starting position of the ore-miner truck is the northwest corner of the field. It must move to the eastern or southern adjacent square, while it can not move to the northern or western adjacent square. And some squares have magic power that can instantaneously transfer the truck to a certain square specified. However, as the captain of the ore-miner truck, you can decide whether to use this magic power or to stay still. One magic power square will never lose its magic power; you can use the magic power whenever you get there.
Input
The first line of the input is an integer T which indicates the number of test cases.
For each of the test case, the first will be two integers N, M (2 ≤ N, M ≤ 40).
The next N lines will describe the map of the mine field. Each of the N lines will be a string that contains M characters. Each character will be an integer X (0 ≤ X ≤ 9) or a '*' or a '#'. The integer X indicates that square hasX units of ores, which your truck could get them all. The '*' indicates this square has a magic power which can transfer truck within an instant. The '#' indicates this square is full of rock and the truck can't move on this square. You can assume that the starting position of the truck will never be a '#' square.
As the map indicates, there are K '*' on the map. Then there follows K lines after the map. The next K lines describe the specified target coordinates for the squares with '*', in the order from north to south then west to east. (the original point is the northwest corner, the coordinate is formatted as north-south, west-east, all from 0 to N - 1,M - 1).
Output
For each test case output the maximum units of ores you can take.
Sample Input
12 2111*0 0
Sample Output
3
拦着我，我要跳馨月湖！
！。
题意：有一个N*M的地图。地图中有三种字符。假设是数字X(0~9)，则表示该区域有X的矿物。假设是"*"，则表示该区域能够传送到某一个确定位置。
假设是"#"，则表示该区域不能进入。如今出发点在地图左上方即(0，0)点，题目保证起点一定不是"#"区域。
要求每次仅仅能向右走、向下走，当遇到传送点的时候能够选择传送到指定位置或者不传送。'*'位置的传送效果能够无限使用，问你从起点出发最多能採多少矿。
思路：
1，把N*M个位置虚拟成N*M个点。依据是否可达的关系建边 来构图。
2，对构成的图求SCC。并统计全部SCC里面的矿物数；
3，缩点后，构建新图。以起点所在的SCC为源点。SPFA求最长路。
注意：
1，'*'的传送位置可能是'#'（题目说了传送位置是合法的。所以传送位置不用考虑是否越界）。并且对于'*'位置，能够选择传送或者不传送。所以必需要向下、右建边（需要考虑越界）。
2。The next K lines describe the specified target coordinates for the squares with '*', in the order from north to south then west to east. 对于出现的'*'。先逐行遍历，再逐列遍历。
好想的思路，好写的代码。不好理解的——两个实现过程为什么一个WA，一个AC。
不知道这两个实现最长路的建图 有什么不同，我算是无语了。 O__O "… 明天再好好想吧，今天有点不舒服。。
。
希望路过的大牛指点迷津。
代码一：
WA了， 我只是设了一个超级源点连通起点所在的SCC，边权为当前SCC的矿物数。
然后缩点建图时边权为终点SCC的矿物数。这样跑SPFA就WA。 曾经写POJ 3160用这样的写法就AC了。
struct Node//用于跑SPFA{ int from, to, val, next;};Node node[MAXM];int Head[MAXN], nodenum;int val[MAXN];//记录每一个SCC的 矿物数void addNode(int u, int v, int w){ Node E = {u, v, w, Head[u]}; node[nodenum] = E; Head[u] = nodenum++;}void suodian(){ nodenum = 0; memset(Head, -1, sizeof(Head)); memset(val, 0, sizeof(val)); //求出每一个SCC里面全部的矿物数 for(int i = 1; i <= scc_cnt; i++) { int sum = 0; for(int j = 0; j < scc[i].size(); j++) sum += num[scc[i][j]]; val[i] = sum; } //保证从左上角出发，虚拟源点0 连通原图中起点所在的SCC 边权为当前SCC的矿物数 addNode(0, sccno[0], val[sccno[0]]); for(int i = 0; i < edgenum; i++) { int u = sccno[edge[i].from]; int v = sccno[edge[i].to]; if(u != v) addNode(u, v, val[v]);//建边 }}int dist[MAXN];bool vis[MAXN];void SPFA(int sx)//跑一遍SPFA 求最长路{ queue Q; memset(dist, 0, sizeof(dist)); memset(vis, false, sizeof(vis)); vis[sx] = true; Q.push(sx); while(!Q.empty()) { int u = Q.front(); Q.pop(); vis[u] = false; for(int i = Head[u]; i != -1; i = node[i].next) { Node E = node[i]; if(dist[E.to] < dist[u] + E.val) { dist[E.to] = dist[u] + E.val; if(!vis[E.to]) { vis[E.to] = true; Q.push(E.to); } } } } sort(dist, dist+scc_cnt+1); printf("%d\n", dist[scc_cnt]);} 代码二：
AC代码的 —— 仅仅建边 不设置边权，用数组记录SCC的矿物数，借此来更新。
struct Node//用于跑SPFA{ int from, to, next;};Node node[MAXM];int Head[MAXN], nodenum;int val[MAXN];//记录每一个SCC的 矿物数void addNode(int u, int v){ Node E = {u, v,Head[u]}; node[nodenum] = E; Head[u] = nodenum++;}void suodian(){ nodenum = 0; memset(Head, -1, sizeof(Head)); memset(val, 0, sizeof(val)); //求出每一个SCC里面全部的矿物数 for(int i = 1; i <= scc_cnt; i++) { int sum = 0; for(int j = 0; j < scc[i].size(); j++) sum += num[scc[i][j]]; val[i] = sum; } for(int i = 0; i < edgenum; i++) { int u = sccno[edge[i].from]; int v = sccno[edge[i].to]; if(u != v) addNode(u, v);//建边 }}int dist[MAXN];bool vis[MAXN];void SPFA()//跑一遍SPFA 求最长路{ queue Q; memset(dist, 0, sizeof(dist)); memset(vis, false, sizeof(vis)); vis[sccno[0]] = true; dist[sccno[0]] = val[sccno[0]]; Q.push(sccno[0]); while(!Q.empty()) { int u = Q.front(); Q.pop(); vis[u] = false; for(int i = Head[u]; i != -1; i = node[i].next) { Node E = node[i]; if(dist[E.to] < dist[u] + val[E.to]) { dist[E.to] = dist[u] + val[E.to]; if(!vis[E.to]) { vis[E.to] = true; Q.push(E.to); } } } } sort(dist, dist+scc_cnt+1); printf("%d\n", dist[scc_cnt]);} 没想通o(╯□╰)o
AC代码：0ms
#include#include #include #include #include #include #define MAXN 2000+10#define MAXM 3000000+10#define INF 0x3f3f3fusing namespace std;struct Edge{ int from, to, next;};Edge edge[MAXM];int head[MAXN], edgenum;int low[MAXN], dfn[MAXN];int dfs_clock;int sccno[MAXN], scc_cnt;stack S;vector scc[MAXN];bool Instack[MAXN];int N, M;void init(){ edgenum = 0; memset(head, -1, sizeof(head));}int point(int x, int y){ return x * M + y;//这里写反了 RE2次。。。
} void addEdge(int u, int v) { Edge E = {u, v, head[u]}; edge[edgenum] = E; head[u] = edgenum++; } bool judge(int x, int y)//推断越界 写多了。。。 { return x >= 0 && x < N && y >= 0 && y < M; } int num[MAXN];//存储相应位置的 矿物数目 void getMap() { scanf("%d%d", &N, &M); int move[2][2] = {0,1, 1,0}; char str[50][50]; int x[MAXN], y[MAXN]; memset(num, 0, sizeof(num)); int k = 0; for(int i = 0; i < N; i++) { scanf("%s", str[i]); for(int j = 0; j < M; j++) if(str[i][j] == '*') k++; } for(int i = 0; i < k; i++) scanf("%d%d", &x[i], &y[i]); k = 0; for(int i = 0; i < N; i++)//注意这里 先遍历行 再遍历列 { for(int j = 0; j < M; j++) { if(str[i][j] == '#')//不能走 continue; for(int p = 0; p < 2; p++)//两个方向 { int a = i + move[p][0]; int b = j + move[p][1]; if(judge(a, b) && str[a][b] != '#')//能走且不能越界 连通 addEdge(point(i, j), point(a, b)); } if(str[i][j] != '*')//数字 有矿物 num[point(i, j)] = str[i][j] - '0';//记录矿物数 else//连通 传送位置 { if(str[x[k]][y[k]] != '#') addEdge(point(i, j), point(x[k], y[k]));//由于这里[k++] WA N次 k++; } } } } void tarjan(int u, int fa) { int v; low[u] = dfn[u] = ++dfs_clock; S.push(u); Instack[u] = true; for(int i = head[u]; i != -1; i = edge[i].next) { v = edge[i].to; if(!dfn[v]) { tarjan(v, u); low[u] = min(low[u], low[v]); } else if(Instack[v]) low[u] = min(low[u], dfn[v]); } if(low[u] == dfn[u]) { scc_cnt++; scc[scc_cnt].clear(); for(;;) { v = S.top(); S.pop(); Instack[v] = false; sccno[v] = scc_cnt; sumval[scc_cnt] += num[v]; scc[scc_cnt].push_back(v); if(v == u) break; } } } void find_cut(int l, int r) { memset(low, 0, sizeof(low)); memset(dfn, 0, sizeof(dfn)); memset(sccno, 0, sizeof(sccno)); memset(sumval, 0, sizeof(sumval)); memset(Instack, false, sizeof(Instack)); dfs_clock = scc_cnt = 0; for(int i = l; i <= r; i++) if(!dfn[i]) tarjan(i, -1); } struct Node//用于跑SPFA { int from, to, next; }; Node node[MAXM]; int Head[MAXN], nodenum; int val[MAXN];//记录每一个SCC的 矿物数 void addNode(int u, int v) { Node E = {u, v,Head[u]}; node[nodenum] = E; Head[u] = nodenum++; } void suodian() { nodenum = 0; memset(Head, -1, sizeof(Head)); memset(val, 0, sizeof(val)); //求出每一个SCC里面全部的矿物数 for(int i = 1; i <= scc_cnt; i++) { int sum = 0; for(int j = 0; j < scc[i].size(); j++) sum += num[scc[i][j]]; val[i] = sum; } //为什么这样写 会WA？？？ 坑死我了 //保证从左上角出发。虚拟源点0 连通原图中起点所在的SCC 边权为当前SCC的矿物数 //addNode(0, sccno[0], val[sccno[0]]); for(int i = 0; i < edgenum; i++) { int u = sccno[edge[i].from]; int v = sccno[edge[i].to]; if(u != v) addNode(u, v);//建边 } } int dist[MAXN]; bool vis[MAXN]; void SPFA()//跑一遍SPFA 求最长路 { queue<int> Q; memset(dist, 0, sizeof(dist)); memset(vis, false, sizeof(vis)); vis[sccno[0]] = true; dist[sccno[0]] = val[sccno[0]]; Q.push(sccno[0]); while(!Q.empty()) { int u = Q.front(); Q.pop(); vis[u] = false; for(int i = Head[u]; i != -1; i = node[i].next) { Node E = node[i]; if(dist[E.to] < dist[u] + val[E.to]) { dist[E.to] = dist[u] + val[E.to]; if(!vis[E.to]) { vis[E.to] = true; Q.push(E.to); } } } } sort(dist, dist+scc_cnt+1); printf("%d\n", dist[scc_cnt]); } void solve() { find_cut(0, N*M-1); suodian(); SPFA(); } int main() { int t; scanf("%d", &t); while(t--) { init(); getMap(); solve(); } return 0; }