[POJ 1179] Polygon【区间DP】
Problem:
| Time Limit: 1000MS | Memory Limit: 10000K |
Description
Polygon is a game for one player that starts on a polygon with N vertices, like the one in Figure 1, where N=4. Each vertex is labelled with an integer and each edge is labelled with either the symbol + (addition) or the symbol * (product). The edges are numbered from 1 to N.
�pick an edge E and the two vertices V1 and V2 that are linked by E; and
�replace them by a new vertex, labelled with the result of performing the operation indicated in E on the labels of V1 and V2.
The game ends when there are no more edges, and its score is the label of the single vertex remaining.Consider the polygon of Figure 1. The player started by removing edge 3. After that, the player picked edge 1, then edge 4, and, finally, edge 2. The score is 0.
�pick an edge E and the two vertices V1 and V2 that are linked by E; and
�replace them by a new vertex, labelled with the result of performing the operation indicated in E on the labels of V1 and V2.
The game ends when there are no more edges, and its score is the label of the single vertex remaining.Consider the polygon of Figure 1. The player started by removing edge 3. After that, the player picked edge 1, then edge 4, and, finally, edge 2. The score is 0.
Input
Your program is to read from standard input. The input describes a polygon with N vertices. It contains two lines. On the first line is the number N. The second line contains the labels of edges 1, ..., N, interleaved with the vertices' labels (first that of the vertex between edges 1 and 2, then that of the vertex between edges 2 and 3, and so on, until that of the vertex between edges N and 1), all separated by one space. An edge label is either the letter t (representing +) or the letter x (representing *).
3 <= N <= 50
For any sequence of moves, vertex labels are in the range [-32768,32767].
3 <= N <= 50
For any sequence of moves, vertex labels are in the range [-32768,32767].
Output
Your program is to write to standard output. On the first line your program must write the highest score one can get for the input polygon. On the second line it must write the list of all edges that, if removed on the first move, can lead to a game with that score. Edges must be written in increasing order, separated by one space.
Sample Input
4 t -7 t 4 x 2 x 5
Sample Output
33 1 2
Source
IOI 1998
Solution:
本来是要练四边形不等式优化来着。。找了道和“石子归并”差不多的题，结果竟然用不了。。
这道题是石子归并的升级版，操作变成了 + 和 * 两种。
直接拆环成链，用区间 DP 套就可以了。。
唯一坑点是由于负数的存在，最大值还可能由 2 个最小值或 1 个最大值和 1 个最小值相乘而得。
所以最小值也要记录。1A ~~~
Code: O(N3)
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
inline char getop(){
char ch;
while((ch = getchar()) != EOF && ch != 't' && ch != 'x');
return ch;
}
int N, v[110];
bool op[110];
int u[110][110], l[110][110];
int main(){
scanf("%d", &N);
for(register int i = 1; i <= N; i++){
op[i + N] = op[i] = (getop() == 't') ? 0 : 1;
scanf("%d", v + i), v[i + N] = v[i];
}
memset(u, 0xc0, sizeof(u)), memset(l, 0x3f, sizeof(l));
for(register int i = 1; i < (N << 1); i++) u[i][i] = l[i][i] = v[i];
for(register int gp = 1; gp < N; gp++)
for(register int i = 1; i < (N << 1) - gp; i++){
int j = i + gp;
for(register int k = i; k < j; k++){
if(op[k + 1]){
u[i][j] = max(u[i][j], max(u[i][k] * u[k + 1][j], max(u[i][k] * l[k + 1][j], max(l[i][k] * u[k + 1][j], l[i][k] * l[k + 1][j]))));
l[i][j] = min(l[i][j], min(u[i][k] * u[k + 1][j], min(u[i][k] * l[k + 1][j], min(l[i][k] * u[k + 1][j], l[i][k] * l[k + 1][j]))));
}
else{
u[i][j] = max(u[i][j], u[i][k] + u[k + 1][j]);
l[i][j] = min(l[i][j], l[i][k] + l[k + 1][j]);
}
}
}
int maxv = 0xc0c0c0c0;
for(register int i = 1; i <= N; i++) maxv = max(maxv, u[i][i + N - 1]);
printf("%d\n", maxv);
bool isfirst = 1;
for(register int i = 1; i <= N; i++)
if(u[i][i + N - 1] == maxv){
if(!isfirst) printf(" %d", i);
else isfirst = 0, printf("%d", i);
}
return 0;
}
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